Question

A 1.555-g sample of baking soda decomposes with heat to produce 0.991 g Na2CO3. Refer to Example Exercise 14.l and show the calculation for the theoretical yield of Na2CO3.
What is the percent yield of sodium carbonate, Na2CO3?
6. A 1473-g unknown mixture with baking soda is heated and has a mass loss of 0.325 g. Refer to Example Exercise 14.2 and show the calculation for the percentage NaHCOs in the mixture.

Answer 1

Answer:

a) 101%

b)59.7%

Explanation:

The equation for the thermal decomposition of baking soda is shown;

2NaHCO3 → Na2CO3 + H2O + CO2

Number of moles of baking soda= mass/molar mass= 1.555g/84.007 g/mol = 0.0185 moles

From the reaction equation;

2 moles of baking soda yields 1 mole of sodium carbonate

0.0185 moles of baking soda will yield = 0.0185 moles ×1 /2 = 9.25 ×10^-3 moles of sodium carbonate.

Therefore, mass of sodium carbonate= 9.25 ×10^-3 moles × 106gmol-1= 0.9805 g of sodium carbonate. This is the theoretical yield of sodium carbonate.

%yield = actual yield/theoretical yield ×100

% yield = 0.991/0.9805 ×100

%yield = 101%

Since ;

2NaHCO3 → Na2CO3 + H2O + CO2

And H2O + CO2 ---> H2CO3

Hence I can write, 2NaHCO3 → Na2CO3 + H2CO3

Molar mass of H2CO3= 62.03 gmol-1

Molar mass of baking soda= 84 gmol-1

Therefore, mass of baking soda=

0.325/62.03 × 2 × 84 = 0.88 g of NaHCO3

% of NaHCO3= 0.88/1.473 × 100 = 59.7%

Answer 2

The decomposition reaction of baking soda is a reaction in which water and carbon dioxide ae given off as gaseous products.

• 5. The theoretical yield of Na₂CO₃ is approximately 0.9809 grams

• The percentage yield of sodium carbonate is approximately 101.02%.

• 6. Percentage of NaHCO₃ in the mixture is approximately 59.76%.

Reasons:

Mass of baking soda = 1.555 g

Mass of Na₂CO₃ produced = 0.991 g

Required:

Calculation for the theoretical yield

Solution:

Theoretical yield (mass) of Na₂CO₃ produced is found as follows;

Molar mass of Na₂CO₃ = 105.9888 g/mol

Molar mass of NaHCO₃ = 84.007 g/mol

$\displaystyle 1.555 \, g \, NaHCO_3 \times \frac{1 \, mol \, NaHCO_3}{84.007 \, g \, NaHCO_3} \times \frac{1 \, mol \, Na_2CO_3}{2 \, mol \, NaHCO_3} \times 105.9888 \ g \approx 0.9809 \, g \, Na_2CO_3$

The theoretical yield of Na₂CO₃ ≈ 0.9809 grams.

The percentage yield is given as follows;

$\displaystyle Percentage \ yield = \mathbf{\frac{Actual \, Yield}{Theorectical \, Yield} \times 100 \%}$

The percentage yield of Na₂CO₃ is therefore;

$\displaystyle Percentage \ yield \ of \ Na_2CO_3= \frac{0.991}{0.9809} \times 100 \% \approx \underline{ 101.02 \%}$

(Some baking soda may remain if the reaction is not completed)

6. Mass of the unknown mixture of baking soda = 1473 g

Mass loss from the mixture = 0.325 g

Required:

The percentage of NaHCO₃ in the mixture.

Solution:

The chemical in the mass loss from heating the NaHCO₃ = H₂CO₃

Molar mass of H₂CO₃ = 62.03 g/mol

$\displaystyle \mathrm{Number \ of \ moles \ of \ H_2CO_3 \ produced} = \frac{0.325 \, g}{62.03 \, g/mol} \approx 5.2394 \times 10^{-3} \ moles$

The chemical reaction is presented as follows;

• 2NaHCO₃(s) $\underrightarrow {\Delta \ Heated}$ Na₂CO₃(s) + H₂CO₃(g)

• 2 moles of NaHCO₃  produces 1 mole of H₂CO₃

The number of moles of NaHCO₃ in the mixture is therefore;

• 2 × 5.2394 × 10⁻³ moles ≈ 1.04788 × 10⁻² moles

Mass of NaHCO₃ in the mixture is therefore

• Mass of NaHCO₃ = 1.04788 × 10⁻²  moles × 84.007 g/mol = 0.88029 g

$\displaystyle Percentage \ of \ NaHCO_3 \ in \ the \ mixture \ = \mathbf{ \frac{Mass \ of \ NaHCO_3}{Mass \ of \ mixture} \times 100}$

Which gives;

• $\displaystyle Percentage \ of \ NaHCO_3 \ in \ the \ mixture \ = \ \frac{0.88029 \, g}{1.473 \, g} \times 100 \approx \underline{ 59.76 \%}$

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