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}=(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* \text{Molar mass of }CO_2`

`\text{Mass of }CO_2=(0.0131mole)* (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}* 100=(0.53g)/(0.5764g)* 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