Question

A sample of 0.53 g of carbon dioxide was obtained by heating 1.31 g of calcium carbonate. what is the percent yield for this reaction? caco3(s) ⟶ cao(s) + co2(s)

Answer 1

Answer : The percent yield for this reaction is, 91.9 %

Solution : Given,

Mass of carbon dioxide = 0.53 g

Mass of calcium carbonate = 1.31 g

Molar mass of carbon dioxide = 44 g/mole

Molar mass of calcium carbonate = 100 g/mole

First we have to calculate the moles of $CaCO_3$.

$\text{Moles of }CaCO_3=\frac{\text{Mass of }CaCO_3}{\text{Molar mass of }CaCO_3}=\frac{1.31g}{100g/mole}=0.0131moles$

Now we have to calculate the mass of carbon dioxide.

The balanced chemical reaction is,

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

From the balanced reaction we conclude that

As, 1 mole of $CaCO_3$ react to give 1 mole of $CO_2$

So, 0.0131 mole of $CaCO_3$ react to give 0.0131 mole of $CO_2$

Now we have to calculate the mass of carbon dioxide.

$\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2$

$\text{Mass of }CO_2=(0.0131mole)\times (44g/mole)=0.5764g$

Now we have to calculate the percent yield for this reaction.

$\%\text{ yield of }CO_2=\frac{\text{Actual yield of }CO_2}{\text{Theoretical yield of }CO_2}\times 100=\frac{0.53g}{0.5764g}\times 100=91.9\%$

Therefore, the percent yield for this reaction is, 91.9 %

Answer 2

CaCO3(s) ⟶ CaO(s)+CO2(s) 

moles CaCO3: 1.31 g/100 g/mole CaCO3= 0.0131 

From stoichiometry, 1 mole of CO2 is formed per 1 mole CaCO3, therefore 0.0131 moles CO2 should also be formed. 
0.0131 moles CO2 x 44 g/mole CO2 = 0.576 g CO2 

Therefore:
% Yield: 0.53/.576 x100= 92 percent yield