Question

Calculate the mass of oxygen produced when 2.45 g of $K{O}_2$ and 4.44 g of $C{O}_2$ react.
$K{O}_2+C{O}_2\to K_{2}C{O}_3+O_2$ (yield = 80 %)
(Molar mass of K = 39 g/mol)

• A. 1.11 g
• B. 0.14 g
• C. 2.05 g
• D. 0.65 g

Answer 1

The correct option is D 0.65 g

$4K{O}_2+2C{O}_2\to 2K_{2}C{O}_3+3O_2$

$\text{Moles of}K{O}_2=\frac{\text{given mass}}{\text{molar mass}}=\frac{2.45}{71}=0.034mol\text{Moles of}C{O}_2=\frac{\text{given mass}}{\text{Molar mass}}=\frac{4.44}{44}=0.100mol$

Finding the limiting reagent:
$ForK{O}_2=\frac{\text{given moles}}{\text{stoichiometric coefficient}}=0.0085molForC{O}_2=\frac{0.1}{2}=0.05mol$
So, $K{O}_2$ is the limiting reagent.
4 moles of $K{O}_2$ produces 3 moles of $O_2$
$\text{0.034 mol will form =}\frac{3}{4}\times 0.034=0.0255mol$

$\text{Mass of oxygen formed =}0.0255mol\times 32g/mol=0.816g$
But, the yield of the reaction is 80%

Actual amount of oxygen formed = $0.816g\times \frac{80}{100}=0.65g$