Ans: Assume 100 g of the compound:

This means we have, 65.45 g C, 5.45 g H, and 29.09 g O.

Convert masses to moles:

Moles of C: $65.45\,\mathrm{g}\times\frac{1\,\mathrm{mol}}{12.01\,\mathrm{g}}=5.45\,\mathrm{mol}$

Moles of H: $5.45\,\mathrm{g}\times\frac{1\,\mathrm{mol}}{1.01\,\mathrm{g}}=5.40\,\mathrm{mol}$

Moles of O: $29.09\,\mathrm{g}\times\frac{1\,\mathrm{mol}}{16.00\,\mathrm{g}}=1.82\,\mathrm{mol}$

Determine the empirical formula:

Divide each mole value by the smallest mole value (1.82 mol) to find the simplest whole-number ratio:

C: $\frac{5.45}{1.82}=3.00$

H: $\frac{5.40}{1.82}=2.97=3$

O: $\frac{1.82}{1.82}=1$

The empirical formula is therefore C₃H₃O.

Calculate the empirical formula mass C₃H₃O:

$(3\times12.01)+(3\times1.01)+(1\times16.00)$

$=36.03+3.03+16.00=55.06\,\mathrm{g/mol}$

Determine the molecular formula:

Divide the molar mass of the compound by the empirical formula mass:

$\frac{110\,\mathrm{g/mol}}{55.06\,\mathrm{g/mol}}=1.99=2$

Multiply the subscripts in the empirical formula by this factor:

$(\mathrm{C_3H_3O})_2=\mathrm{C_6H_6O_2}$ Therefore, the molecular formula of the compound is C₆H₆O₂.

Reaction: $\mathrm{2AlCl_3(s)+3H_2O(l)\longrightarrow Al_2O_3(s)+6HCl(aq)}$

Ans: Given Information:

Mass of AlCl₃ = 125g

Reaction: $\mathrm{2AlCl_3(s)+3H_2O(l)\longrightarrow Al_2O_3(s)+6HCl(aq)}$

Goal:

• Find the number of moles of HCl produced.

Steps:

Calculate the molar mass of AlCl₃:

Al: 26.98 g/mol
Cl: 35.45 g/mol

$M_{\mathrm{AlCl_3}}=26.98+(3\times35.45)=26.98+106.35=133.33\ \mathrm{g/mol}$

Convert grams of AlCl₃ to moles:

$n_{\mathrm{AlCl_3}}=\frac{125\ \mathrm{g}}{133.33\ \mathrm{g/mol}}=0.937\ \mathrm{mol}$

Use the stoichiometric ratio from the balanced equation to find moles of HCl:

From the balanced equation, 2 moles of AlCl₃ produce 6 moles of HCl.

So, the ratio is $\frac{6\ \mathrm{mol\ HCl}}{2\ \mathrm{mol\ AlCl_3}}$, or 3 mol HCl : 1 mol AlCl₃.

$n_{\mathrm{HCl}}=0.937\ \mathrm{mol\ AlCl_3}\times\frac{6\ \mathrm{mol\ HCl}}{2\ \mathrm{mol\ AlCl_3}}$

$=0.937\ \mathrm{mol\ AlCl_3}\times3=2.811\ \mathrm{mol\ HCl}$

Answer:

Therefore, 2.811 moles of HCl will be produced when 125 g of AlCl₃ reacts with excess water.

$n_{\mathrm{HCl}}=2.811\ \mathrm{mol}$

Reaction: $\mathrm{Ca(s)+3O_2(g)+N_2(g)\longrightarrow Ca(NO_3)_2(s)}$

Ans: Given Information:

Mass of Ca(NO₃)₂ = 1.0 g

Reaction: $\mathrm{Ca(s)+3O_2(g)+N_2(g)\longrightarrow Ca(NO_3)_2(s)}$

Goal:
Find the number of moles of O₂ needed.

Steps:

Calculate the molar mass of Ca(NO₃)₂:

• Ca: 40.08 g/mol
• N: 14.01 g/mol
• O: 16.00 g/mol
• $M_{\mathrm{Ca(NO_3)_2}}=40.08+2\times(14.01+3\times16.00)$
  $=40.08+2\times(14.01+48.00)$
  $=40.08+2\times62.01=40.08+124.02=164.10\ \mathrm{g/mol}$

• Convert grams of Ca(NO₃)₂ to moles:

  $n_{\mathrm{Ca(NO_3)_2}}=\frac{1.0\ \mathrm{g}}{164.10\ \mathrm{g/mol}}=0.00609\ \mathrm{mol}$

Use the stoichiometric ratio from the balanced equation to find moles of O₂:

From the balanced equation, 3 moles of O₂ produce 1 mole of Ca(NO₃)₂.

So, the ratio is $\frac{3\ \mathrm{mol\ O_2}}{1\ \mathrm{mol\ Ca(NO_3)_2}}$.

$n_{\mathrm{O_2}}=0.00609\ \mathrm{mol\ Ca(NO_3)_2}\times\frac{3\ \mathrm{mol\ O_2}}{1\ \mathrm{mol\ Ca(NO_3)_2}}=0.01827\ \mathrm{mol\ O_2}$

Answer:

• Therefore, 0.01827 moles of O₂ are needed to produce 1.0 g of Ca(NO₃)₂.
• $n_{\mathrm{O_2}}=0.01827\ \mathrm{mol}$

Reaction: $\mathrm{2Al(s)+3S(s)\longrightarrow Al_2S_3(s)}$

Ans: 1. Given Information:

Reaction: $\mathrm{2Al(s)+3S(s)\longrightarrow Al_2S_3(s)}$

Mass of Al = 25g
Mass of S = 25g

Goal:

• Determine the limiting reactant.

Steps:

Calculate the molar masses of Al and S:

Al: 26.98 g/mol
S: 32.07 g/mol

• Convert grams of Al and S to moles:

  $n_{\mathrm{Al}}=\frac{25\ \mathrm{g}}{26.98\ \mathrm{g/mol}}=0.926\ \mathrm{mol}$
  $n_{\mathrm{S}}=\frac{25\ \mathrm{g}}{32.07\ \mathrm{g/mol}}=0.779\ \mathrm{mol}$
• Use the stoichiometric ratio from the balanced equation to determine the limiting reactant:
  From the balanced equation, 2 moles of Al react with 3 moles of S.
• Calculate how many moles of S are needed to react with the available Al:
  Moles of S needed = $0.926\ \mathrm{mol\ Al}\times\frac{3\ \mathrm{mol\ S}}{2\ \mathrm{mol\ Al}}=1.389\ \mathrm{mol\ S}$
• Calculate how many moles of Al are needed to react with the available S:
  Moles of Al needed = $0.779\ \mathrm{mol\ S}\times\frac{2\ \mathrm{mol\ Al}}{3\ \mathrm{mol\ S}}=0.519\ \mathrm{mol\ Al}$
• Compare the required moles with the available moles:
  We have 0.926 mol of Al, and we only need 0.519 mol of Al to react with all the S. Therefore, we have enough Al.

We have 0.779 mol of S, but we need 1.389 moles of S to react with all the Al. Therefore, we don't have enough S.

Answer: Since we don't have enough sulfur to react with all of the aluminum, sulfur (S) is the limiting reactant.

Ans: Assume 100 g of the compound:

This means we have 20.2 g Na, 37.6 g S, and 42.2 g O.

Convert masses to moles:

Moles of Na: $20.2\,\mathrm{g}\times\frac{1\,\mathrm{mol}}{22.99\,\mathrm{g}}=0.879\,\mathrm{mol}$

Moles of S: $37.6\,\mathrm{g}\times\frac{1\,\mathrm{mol}}{32.07\,\mathrm{g}}=1.172\,\mathrm{mol}$

Moles of O: $42.2\,\mathrm{g}\times\frac{1\,\mathrm{mol}}{16.00\,\mathrm{g}}=2.638\,\mathrm{mol}$

Determine the empirical formula:

Divide each mole value by the smallest mole value (0.879 mol) to find the simplest whole-number ratio:

Na: $\frac{0.879}{0.879}=1$

S: $\frac{1.172}{0.879}=1.333\approx\frac{4}{3}$

O: $\frac{2.638}{0.879}=3.001\approx3$

To get whole numbers, multiply all the subscripts by 3:

Na: $1\times3=3$
S: $\frac{4}{3}\times3=4$
O: $3\times3=9$

The empirical formula is therefore Na₃S₄O₉.

Calculate the empirical formula mass:

$(3\times22.99)+(4\times32.07)+(9\times16.00)$

$=68.97+128.28+144.00=341.25\,\mathrm{g/mol}$

Determine the molecular formula:

Divide the molar mass of the compound by the empirical formula mass:

$\frac{682.8\,\mathrm{g/mol}}{341.25\,\mathrm{g/mol}}=2.001\approx2$

Multiply the subscripts in the empirical formula by this factor:

$(\mathrm{Na_3S_4O_9})_2=\mathrm{Na_6S_8O_{18}}$

Therefore, the empirical formula of the compound is Na₃S₄O₉, and the molecular formula is Na₆S₈O₁₈.

Ans: Given Information:

Reaction: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)

Mass of CaCO₃ = 68.1 g
Mass of HCl = 51.6 g

Goal: Determine the limiting reactant.

Steps: Calculate the molar masses of CaCO₃ and HCl:

Ca: 40.08 g/mol
C: 12.01 g/mol
O: 16.00 g/mol
H: 1.01 g/mol
Cl: 35.45 g/mol

$M_{\mathrm{CaCO_3}}=40.08+12.01+(3\times16.00)=40.08+12.01+48.00=100.09\,\mathrm{g/mol}$

$M_{\mathrm{HCl}}=1.01+35.45=36.46\,\mathrm{g/mol}$

Convert grams of CaCO₃ and HCl to moles:

$n_{\mathrm{CaCO_3}}=\frac{68.1\,\mathrm{g}}{100.09\,\mathrm{g/mol}}=0.680\,\mathrm{mol}$

$n_{\mathrm{HCl}}=\frac{51.6\,\mathrm{g}}{36.46\,\mathrm{g/mol}}=1.415\,\mathrm{mol}$

Use the stoichiometric ratio from the balanced equation to determine the limiting reactant:

From the balanced equation, 1 mole of CaCO₃ reacts with 2 moles of HCl.

Calculate how many moles of HCl are needed to react with the available CaCO₃:

Moles of HCl needed = $0.680\,\mathrm{mol\ CaCO_3}\times\frac{2\,\mathrm{mol\ HCl}}{1\,\mathrm{mol\ CaCO_3}}=1.360\,\mathrm{mol\ HCl}$

• Calculate how many moles of CaCO₃ are needed to react with the available HCl:
  Moles of CaCO₃ needed = $1.415\,\mathrm{mol\ HCl}\times\frac{1\,\mathrm{mol\ CaCO_3}}{2\,\mathrm{mol\ HCl}}=0.708\,\mathrm{mol\ CaCO_3}$
• Compare the required moles with the available moles:
  We have 0.680 mol of CaCO₃, and we need 0.708 mol of CaCO₃ to react with all the HCl. Therefore, we don't have enough CaCO₃.
  We have 1.415 mol of HCl, and we only need 1.360 mol of HCl to react with all the CaCO₃. Therefore, we have enough HCl.

Answer: Since we don't have enough calcium carbonate to react with all of the hydrochloric acid, calcium carbonate (CaCO₃) is the limiting reactant.

Reaction: $\mathrm{CuO_{(s)} + H_2SO_4{}_{(aq)} \longrightarrow CuSO_4{}_{(aq)} + H_2O_{(l)}}$

Ans: Given Information:

Actual yield of CuSO₄ = 1.85g
Theoretical yield of CuSO₄ = 2.5g

Goal: Calculate the percentage yield.

Formula: Percentage Yield = $\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%$

Calculation: Percentage Yield = $\frac{1.85\,\mathrm{g}}{2.5\,\mathrm{g}}\times 100\%=0.74\times100\%=74\%$

Answer: The percentage yield of the reaction is 74%.

Reaction: $\mathrm{CaCO_3{}_{(s)} \longrightarrow CaO_{(s)} + CO_2{}_{(g)}}$

Ans: Given Information:

Reaction: CaCO₃(s) → CaO(s) + CO₂(g)

Volume of CO₂ produced = 24 dm³

RTP conditions (assumed): 1 mol of gas occupies 24 dm³

Goal: Find the mass of CaCO₃ heated.

Steps: Calculate moles of CO₂ produced at RTP:

Since 1 mole of any gas occupies 24 dm³ at RTP, then:

$n_{\mathrm{CO_2}}=\frac{24\,\mathrm{dm^3}}{24\,\mathrm{dm^3/mol}}=1\,\mathrm{mol}$

• Use the stoichiometric ratio from the balanced equation to find moles of CaCO₃:

From the balanced equation, 1 mole of CaCO₃ produces 1 mole of CO₂.

So, $n_{\mathrm{CaCO_3}}=n_{\mathrm{CO_2}}=1\,\mathrm{mol}$

Calculate the molar mass of CaCO₃:

Ca: 40.08 g/mol
C: 12.01 g/mol
O: 16.00 g/mol

$M_{\mathrm{CaCO_3}}=40.08+12.01+(3\times16.00)=40.08+12.01+48.00$
$=100.09\,\mathrm{g/mol}$

Convert moles of CaCO₃ to grams:

$m_{\mathrm{CaCO_3}}=1\,\mathrm{mol}\times100.09\,\mathrm{g/mol}=100.09\,\mathrm{g}$

Answer:

• Therefore, 100.09 g of limestone (CaCO₃) was heated.
• $m_{\mathrm{CaCO_3}}=100.09\,\mathrm{g}$

Ans: Given Information:

C₂H₄(g) is the chemical formula for ethylene, also known as ethene.

Reaction: C₂H₄(g) + O₂(g) → CO₂(g) + H₂O(g) (unbalanced)

Moles of C₂H₄ = 0.480 mol
Moles of O₂ = 1.08 mol

Goal: Find the number of moles of CO₂ produced.

Steps: Balance the chemical equation:

C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g)

Use the stoichiometric ratio from the balanced equation to determine the limiting reactant:

From the balanced equation, 1 mole of C₂H₄ reacts with 3 moles of O₂.

Calculate how many moles of O₂ are needed to react with the available C₂H₄:

Moles of O₂ needed = 0.480 mol C₂H₄ × (3 mol O₂ / 1 mol C₂H₄) = 1.44 mol O₂

Calculate how many moles of C₂H₄ are needed to react with the available O₂:

Moles of C₂H₄ needed = 1.08 mol O₂ × (1 mol C₂H₄ / 3 mol O₂) = 0.36 mol C₂H₄

• Compare the required moles with the available moles:

  We have 0.480 mol of C₂H₄, and we only need 0.36 mol of C₂H₄ to react with all the O₂. Therefore, we have enough C₂H₄.

  We have 1.08 mol of O₂, but we need 1.44 mol of O₂ to react with all the C₂H₄. Therefore, we don't have enough O₂.

  Therefore, O₂ is the limiting reactant.

• Use the stoichiometric ratio from the balanced equation and the moles of the limiting reactant to find moles of CO₂ produced:

  From the balanced equation, 3 moles of O₂ produce 2 moles of CO₂.

  Moles of CO₂ produced = 1.08 mol O₂ × (2 mol CO₂ / 3 mol O₂) = 0.72 mol CO₂

Answer: Therefore, 0.72 moles of CO₂ can be produced.

$n_{\mathrm{CO_2}} = 0.72\,\mathrm{mol}$

Which reactant will be left un-reacted?

C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g)

To determine which reactant will be left unreacted, we need to identify the limiting reactant and then calculate how much of the excess reactant will be used up.

Ans: Given Information:

Balanced Reaction:

C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g)

From the previous problem, we know that we have 0.480 mol of C₂H₄ and 1.08 mol of O₂. We also determined that O₂ is the limiting reactant.

Goal: Determine which reactant will be left unreacted and how much of it will remain.

Steps: Calculate how much C₂H₄ is used up by the reaction:

Since O₂ is the limiting reactant, we use its amount (1.08 mol) to determine how much C₂H₄ reacts.

From the balanced equation, 3 moles of O₂ react with 1 mole of C₂H₄.

Moles of C₂H₄ used = 1.08 mol O₂ × (1 mol C₂H₄ / 3 mol O₂) = 0.36 mol C₂H₄

Calculate how much C₂H₄ is left unreacted:

Moles of C₂H₄ initially = 0.480 mol
Moles of C₂H₄ used = 0.36 mol
Moles of C₂H₄ remaining = 0.480 mol − 0.36 mol = 0.12 mol

Answer: C₂H₄ will be left unreacted; 0.12 mol remains.

Ans: Given Information:

Mass of CH₄ = 48 g
Excess O₂
RTP conditions (assumed): 1 mol of gas occupies 24 dm³

Goal: Find the volume of CO₂ produced.

Steps: Write and balance the chemical equation:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

Calculate the molar mass of CH₄:

C: 12.01 g/mol
H: 1.01 g/mol

$M_{\mathrm{CH_4}} = 12.01 + (4 \times 1.01) = 12.01 + 4.04 = 16.05\,\mathrm{g/mol}$

Convert grams of CH₄ to moles:

$n_{\mathrm{CH_4}} = \frac{48\,\mathrm{g}}{16.05\,\mathrm{g/mol}} = 2.991\,\mathrm{mol}$

• Use the stoichiometric ratio from the balanced equation to find moles of CO₂ produced:

  From the balanced equation, 1 mole of CH₄ produces 1 mole of CO₂.

  So, $n_{\mathrm{CO_2}} = n_{\mathrm{CH_4}} = 2.991\,\mathrm{mol}$

• Calculate the volume of CO₂ produced at RTP:

  Since 1 mole of any gas occupies 24 dm³ at RTP, then:

  $V_{\mathrm{CO_2}} = 2.991\,\mathrm{mol} \times 24\,\mathrm{dm^3/mol} = 71.784\,\mathrm{dm^3}$

Answer: Therefore, 71.784 L of CO₂ will be produced at room temperature and pressure.

$V_{\mathrm{CO_2}} = 71.784\,\mathrm{dm^3}$

Ans: Given Information:

• 40% Carbon (C)
• 6.7% Hydrogen (H)
• 53.3% Oxygen (O)

Goal: Determine the empirical formula.

Steps:

• Assume 100 g of the compound:
  This means we have 40 g C, 6.7 g H, and 53.3 g O.
• Convert masses to moles:

Moles of C = $40\,\mathrm{g}\times\frac{1\,\mathrm{mol}}{12.01\,\mathrm{g}}=3.33\,\mathrm{mol}$

Moles of H = $6.7\,\mathrm{g}\times\frac{1\,\mathrm{mol}}{1.01\,\mathrm{g}}=6.63\,\mathrm{mol}$

Moles of O = $53.3\,\mathrm{g}\times\frac{1\,\mathrm{mol}}{16.00\,\mathrm{g}}=3.33\,\mathrm{mol}$

• Determine the empirical formula:
  Divide each mole value by the smallest mole value (3.33 mol) to find the simplest whole-number ratio:

C = $\frac{3.33}{3.33}=1$
H = $\frac{6.63}{3.33}=1.99\approx2$
O = $\frac{3.33}{3.33}=1$

Answer: The empirical formula is therefore CH₂O.

Ans: Given Information:

Mass of NaCl = 10.0 g
Volume of solution = 500 cm³

Goal: Calculate the concentration in g/dm³ and calculate the concentration in mol/dm³.

Steps:

• Convert the volume from cm³ to dm³:
  1 dm³ = 1000 cm³
  Volume in dm³ = $500\,\mathrm{cm^3}\times\frac{1\,\mathrm{dm^3}}{1000\,\mathrm{cm^3}}=0.5\,\mathrm{dm^3}$
• Calculate the concentration in g/dm³:
  Concentration (g/dm³) = Mass of NaCl (g) / Volume of solution (dm³) = $\frac{10.0\,\mathrm{g}}{0.5\,\mathrm{dm^3}}=20\,\mathrm{g/dm^3}$
• Calculate the molar mass of NaCl:
  Na: 22.99 g/mol
  Cl: 35.45 g/mol
  $M_{\mathrm{NaCl}}=22.99+35.45=58.44\,\mathrm{g/mol}$
• Convert grams of NaCl to moles:
  $n_{\mathrm{NaCl}}=\frac{10.0\,\mathrm{g}}{58.44\,\mathrm{g/mol}}=0.171\,\mathrm{mol}$
• Calculate the concentration in mol/dm³ (Molarity):
  Concentration (mol/dm³) = Moles of NaCl (mol) / Volume of solution (dm³)
  $=\frac{0.171\,\mathrm{mol}}{0.5\,\mathrm{dm^3}}=0.342\,\mathrm{mol/dm^3}$

Answer:

• The concentration of the solution is 20 g/dm³.
• The concentration of the solution is 0.342 mol/dm³.

Ans: Given Information:

Volume of NaOH = 25.0 cm³ = 0.025 dm³
Concentration of NaOH = 0.10 mol/dm³
Volume of HCl = 20.0 cm³ = 0.020 dm³

Goal: Calculate the concentration of the HCl solution in mol/dm³.

Balanced Chemical Equation:

NaOH_(aq) + HCl_(aq) → NaCl_(aq) + H₂O_(l)

Steps:

• Calculate moles of NaOH used:
  Moles of NaOH = Concentration of NaOH × Volume of NaOH
  $n_{\mathrm{NaOH}}=0.10\,\mathrm{mol/dm^3}\times0.025\,\mathrm{dm^3}=0.0025\,\mathrm{mol}$
• Use the stoichiometric ratio to find moles of HCl:
  From the balanced equation, 1 mole of NaOH reacts with 1 mole of HCl. Therefore, $n_{\mathrm{HCl}}=n_{\mathrm{NaOH}}=0.0025\,\mathrm{mol}$.
• Calculate the concentration of the HCl solution:
  Concentration of HCl = Moles of HCl / Volume of HCl
  $c_{\mathrm{HCl}}=\frac{0.0025\,\mathrm{mol}}{0.020\,\mathrm{dm^3}}=0.125\,\mathrm{mol/dm^3}$

Answer: The concentration of the HCl solution is 0.125 mol/dm³.
Concentration of HCl = $c_{\mathrm{HCl}}=0.125\,\mathrm{mol/dm^3}$

1. Define concentration in the context of this solution.
2. Calculate the mass of NaCl needed to prepare 250 cm³ of the solution.
3. Convert the concentration of the prepared solution to g/dm³.

d. Propose a step-by-step procedure to accurately prepare this solution.

a. Define concentration in the context of this solution

In this context, concentration refers to the amount of NaCl (solute) present in a given volume of the solution. Specifically, a 0.5 mol/dm³ NaCl solution means that there are 0.5 moles of NaCl dissolved in every 1 dm³ (or 1 liter) of the solution.

b. Calculate the mass of NaCl needed to prepare 250 cm³ of the solution

Convert volume to dm³:
250 cm³ × (1 dm³ / 1000 cm³) = 0.25 dm³

Calculate moles of NaCl needed:
Moles = Concentration × Volume
n_NaCl = 0.5 mol/dm³ × 0.25 dm³ = 0.125 mol

Calculate the molar mass of NaCl:
Na = 22.99 g/mol
Cl = 35.45 g/mol
M_NaCl = 22.99 + 35.45 = 58.44 g/mol

Convert moles to grams:
Mass = Moles × Molar Mass
m_NaCl = 0.125 mol × 58.44 g/mol = 7.305 g

Therefore, you need 7.305 g of NaCl.

c. Convert the concentration of the prepared solution to g/dm³

We know the concentration is 0.5 mol/dm³. We also know the molar mass of NaCl is 58.44 g/mol.

Convert mol/dm³ to g/dm³:
Concentration (g/dm³) = Concentration (mol/dm³) × Molar Mass (g/mol)
Concentration (g/dm³) = 0.5 mol/dm³ × 58.44 g/mol = 29.22 g/dm³

Therefore, the concentration is 29.22 g/dm³.

d. Propose a step-by-step procedure to accurately prepare this solution

1. Weigh the NaCl: Accurately weigh out 7.305 g of solid NaCl using an analytical balance.
2. Dissolve the NaCl: Transfer the weighed NaCl to a clean 250 cm³ volumetric flask. Add distilled water to the flask until the NaCl is completely dissolved. You may need to swirl the flask gently to aid dissolution.
3. Fill to the mark: Carefully add more distilled water to the flask until the solution reaches the 250 cm³ mark on the neck of the flask. Ensure that the bottom of the meniscus (the curve of the liquid surface) aligns with the mark at eye level.
4. Mix Thoroughly: Stopper the flask securely. Invert the flask several times (at least 20 times) to ensure the solution is homogeneous and thoroughly mixed.

Answers

1. Concentration is the amount of solute (NaCl) per unit volume of solution. A 0.5 mol/dm³ solution contains 0.5 moles of NaCl in 1 dm³ of solution.
2. 7.305 g of NaCl is needed.
3. The concentration is 29.22 g/dm³.
4. See the step-by-step procedure above.

A critical assessment of why the percentage yield in a reaction is rarely 100% and the factor; contributing to a lower yield:

To double the concentration of a solution, you can either reduce the volume of the solvent by half, double the amount of solute dissolved in it, or dilute a higher concentration solution with the appropriate volume of solvent.

To double the concentration of a solution without changing its volume, you need to add more solute. Here's how you can calculate the amount of solute to add:

1. Determine the desired concentration: You want to double the concentration from 0.5 mol/dm³ to 1.0 mol/dm³.
2. Understand the relationship between concentration, moles, and volume:
   Concentration = Moles of solute / Volume of solution
3. Assume a volume: For simplicity, let's assume you have 1 dm³ of the solution. This means you currently have 0.5 mol of solute in 1 dm³ of solution.
4. Calculate the moles of solute needed for the desired concentration: To achieve a concentration of 1.0 mol/dm³ in 1 dm³ of solution, you need 1.0 mol of solute.
5. Calculate the additional moles of solute needed: You already have 0.5 mol of solute, and you need 1.0 mol. Therefore, you need to add an additional 0.5 mol of solute.

Reaction: $\mathrm{Na_2CO_3 + 2HCl_{(aq)} \longrightarrow 2NaCl_{(aq)} + CO_2{}_{(g)} + H_2O_{(l)}}$

Ans: Write the balanced chemical equation:
$\mathrm{Na_2CO_3 + 2HCl \longrightarrow 2NaCl + CO_2 + H_2O}$

Calculate moles of HCl used:
Volume of HCl = 25.0 cm³ = 0.025 dm³
Concentration of HCl = 0.1 mol/dm³
Moles of HCl = Concentration × Volume
= 0.1 mol/dm³ × 0.025 dm³ = 0.0025 mol

Determine moles of Na₂CO₃ reacted:
From the balanced equation, 1 mole of Na₂CO₃ reacts with 2 moles of HCl.
Moles of Na₂CO₃ = ½ × Moles of HCl
= ½ × 0.0025 mol = 0.00125 mol

Calculate the mass of Na₂CO₃ in the sample:
Molar mass of Na₂CO₃ = (2 × 23) + 12 + (3 × 16)
= 46 + 12 + 48 = 106 g/mol

• Mass of Na₂CO₃ = Moles × Molar mass
  = 0.00125 mol × 106 g/mol = 0.1325 g

Calculate the percentage purity:
Purity = (Mass of Na₂CO₃ / Mass of sample) × 100
Purity = (0.1325 g / 1.00 g) × 100 = 13.25%

Therefore, the percentage purity of the sodium carbonate sample is 13.25%.

Ans: How to convert the concentration of the H₂SO₄ solution from mol/dm³ to g/dm³:

Identify the given concentration:
Concentration = 2 mol/dm³

Determine the molar mass of H₂SO₄:
Molar mass of H₂SO₄ = (2 × 1) + 32 + (4 × 16) = 2 + 32 + 64 = 98 g/mol

Use the relationship between moles, mass, and molar mass:
Mass = Moles × Molar mass

Convert the concentration from mol/dm³ to g/dm³:
Concentration in g/dm³ = Concentration in mol/dm³ × Molar mass
Concentration in g/dm³ = 2 mol/dm³ × 98 g/mol = 196 g/dm³

Therefore, the concentration of the H₂SO₄ solution is 196 g/dm³.

Ans:

$C_1V_1 = C_2V_2$

where C₁ and V₁ are the concentration and volume of acetic acid, and C₂ and V₂ are the concentration and volume of NaOH. Solve for C₁.