Question

The decomposition of 3.32 g CaCO3 yields 1.24 g CaO. What is the percent yield of this reaction? (CaCO3 --> CaO + CO2)

Answer 1

Answer:

66.57%

Explanation:

The decomposition reaction of calcium carbonate is shown below as:

$CaCO_3\rightarrow CaO+CO_2$

Calculation of moles of $CaCO_3$ :

Amount = 3.32 g

Molar mass of $CaCO_3$ = 100 g/mol

The formula for the calculation of moles is shown below:

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

Thus, moles are:

$moles= \frac{3.32\ g}{100\ g/mol}$

$moles= 0.0332\ mol$

According to reaction,

0.0332 moles of $CaCO_3$ decomposes to yield 0.0332 moles of $CaO$

Theoretical yield = 0.0332 moles

Calculation of moles of $CaO$ formed as:

Amount = 1.24 g

Molar mass of $CaO$ = 56 g/mol

The formula for the calculation of moles is shown below:

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

Thus, moles are:

$moles= \frac{1.24\ g}{56\ g/mol}$

$moles= 0.0221\ mol$

Experimental yield = 0.0221 moles

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

Thus,

$Yield\%=\frac {0.0221}{0.0332}\times 100$

Percent yield = 66.57%