Question

Calculate the limiting reagent in the following problem. When 25.0g of P4 reacts with 50.0g of O2, How much P4O10 is formed? P4 + 5 O2=>
P4O10

Answer 1

$P_4 +5O_2\longrightarrow P_4O_{10}$

Here moles of $P_4=\frac{given mass}{molar mass}=\frac{25}{124}=0.201$ mole

moles of $O_2= \frac{given mass}{molar mass}=\frac{50}{32}=1.5625$ mole

here from given reaction 1 mole of P₄ react with 5 mole of O₂

so 1.5625 mole of O₂ react with $=\frac{1}{5}\times1.5625=0.3125$ mole of P₄

So here limiting reagent is P₄

Mole of $P_4O_{10}$ produce from 0.201 mole of P₄= 0.201 mole

So, mass of P₄O₁₀ produce= $mass\times molar mass$ = $0.201\times$ 283.88=56.94 g

Mass of P4O10 produce= 56.94 g