Question

Quicklime, CaO, can be prepared by roasting limestone, CaCO3, according to the following reaction. CaCO3(s) ∆→ CaO(s) + CO2(g). When 2.00 × 103 g CaCO3 is heated, the actual yield of CaO is 1.05 × 103 g. What is the percentage yield?

Answer 1

Answer:

93.71 %

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For $CaCO_3$  :-

Mass of $CaCO_3$  = $2.00\times 10^3$ g

Molar mass of water = 100.0869 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{2.00\times 10^3\ g}{100.0869\ g/mol}$

$Moles\ of\ CaCO_3= 19.98\ mol$

According to the given reaction:

$CaCO_3_{(s)}\rightarrow CaO_{(s)}+CO_2_{(g)}$

1 mole of $CaCO_3$ on reaction forms 1 mole of $CaO$

19.98 mole of $CaCO_3$ on reaction forms 19.98 mole of $CaO$

Moles of $CaO$ = 19.98 moles

Molar mass of $CaO$ = 56.0774 g/mol

Mass of sodium sulfate = Moles × Molar mass = 19.98 × 56.0774 g = 1120.43 g

The expression for the calculation of the percentage yield for a chemical reaction is shown below as:-

$\%\ yield =\frac {Experimental\ yield}{Theoretical\ yield}\times 100$

Given , Values from the question:-

Theoretical yield = 1120.43 g

Experimental yield = 1050 g

Applying the values in the above expression as:-

$\%\ yield =\frac{1050}{1120.43}\times 100$

$\%\ yield =93.71\ \%$