Question

If the percent yield for the following reaction is 65.0%, how many grams of KClO3 are needed to produce 42.0 g of O2? 2 KClO3(s) → 2 KCl(s) + 3 O2(g)

Answer 1

Answer: 163 grams

Explanation: According to avogadro's law, 1 mole of every substance contains avogadro's number $6.023\times 10^{23}$ of particles and weighs equal to the molecular mass.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar Mass}}$

$\text{Number of moles}=\frac{42g}{32g/mol}=1.31moles$

$2KClO_3(s)\rightarrow 2KCl(s)+3O_2(g)$

According to the stochiometry, 3 moles of $O_2$ are produced by 2 moles of $KCO_3$

Thus to produce 1.31 moles of $O_2=[tex]\frac{2}{3}\times 1.31=0.87moles$

But as the percent yield is 65%, the moles of $KClO_3$ required will be $=0.87\times \frac{100}{65}=1.33moles$

Mass of $KClO_3=moles\times {\text {Molar mass}}=1.33\times 122.5=163g$