Question

What mass of hydrogen peroxide will form when 16.9 g of barium peroxide is treated with 8.96 L of HCl gas at STP?
(Molar mass of Ba is 137 g/mol)
$Ba{O}_2\left(s\right)+2HCl\left(g\right)\to BaC{l}_2\left(aq\right)+H_2{O}_2\left(aq\right)$

• A. 2.4 g
• B. 3.4 g
• C. 5.6 g
• D. 7.8 g

Answer 1

The correct option is B 3.4 g

$Ba{O}_2\left(s\right)+2HCl\left(g\right)\to BaC{l}_2\left(aq\right)+H_2{O}_2\left(aq\right)$
$\text{Moles of}Ba{O}_2=\frac{\text{given mass}}{\text{molar mass}}=\frac{16.9}{169}=0.1mol\text{Moles of HCl}=\frac{\text{given volume at STP}}{\text{molar volume at STP}}=\frac{8.96}{22.4}=0.4mol$

Finding the limiting reagent:
$\text{For}Ba{O}_2=\frac{\text{given moles}}{\text{stoichiometric coefficient}}=0.1\text{for HCl}=\frac{0.4}{2}=0.2$
So, $Ba{O}_2$ is the limiting reagent
1 mol of $Ba{O}_2$ produces 1 mol of $H_2{O}_2$
0.1 mol will produce 0.1 mol of $H_2{O}_2$
Amount of $H_2{O}_2$ produced $=0.1\times 34=3.4$ g