Question

27.6 g of $K_{2}C{O}_3$ is thermally decomposed to give $K_{2}O$ and $C{O}_2$(g). Calculate the volume of $C{O}_2$ formed at STP. (Molar mass of $K_{2}C{O}_3=138gmo{l}^{-1})$

• A. 11.20 L
• B. 4.48 L
• C. 3.20 L
• D. 5.56 L

Answer 1

The correct option is B 4.48 L

$K_{2}C{O}_3\left(s\right)\to K_{2}O\left(s\right)+C{O}_2\left(g\right)$

$\text{Applying the principle of atom conservation (POAC) to C,}\text{Moles of C in}{K}_{2}C{O}_3=\text{moles of carbon in}C{O}_{2}1\times \text{Number of mol of}{K}_{2}C{O}_3=1\times \text{number of moles of}C{O}_2$
Hence, the number of moles of $C{O}_2$ produced will be equal to the number of moles of $K_{2}C{O}_3\text{decomposed.}$
Now we have,
$\text{Moles of}{K}_{2}C{O}_3=\frac{Givenweight}{MolarMass}=\frac{27.6}{138}=0.2$
$\text{Moles of}{K}_{2}C{O}_3=\text{Moles of}C{O}_2=0.2mol\text{We know that the volume of 1 mol}\text{of gas at STP is 22.4 L and}\text{Volume of}C{O}_2=\text{number of mol of}C{O}_2\times 22.4$= 0.2×22.4= 4.48 L
$\text{Therefore, the volume of}C{O}_2=4.48L$