Question

A 13 g sample of P4010 contains how many

Answer 1

Answer:

$\large \boxed{5.5 \times 10^{22}\text{ molecules of P _{2} O}_{5}}$

Explanation:

You must calculate the moles of P₄O₁₀, convert to moles of P₂O₅,  then convert to molecules of P₂O₅.

1. Moles of P₄O₁₀

$\text{Moles of P _{4} O}_{10} = \text{13 g P _{4} O}_{10} \times \dfrac{\text{1 mol P _{4} O}_{10}}{\text{283.89 g P _{4} O}_{10}} = \text{0.0458 mol P _{4} O}_{10}$

2. Moles of P₂O₅

P₄O₁₀ ⟶ 2P₂O₅

The molar ratio is 2 mol P₂O₅:1 mol P₄O₁₀

$\text{Moles of P _{2} O}_{5} = \text{0.0458 mol P _{4} O}_{10} \times \dfrac{\text{2 mol P _{2} O}_{5}}{\text{1 mol P _{4} O}_{10}} = \text{0.0916 mol P _{2} O}_{5}$

3. Molecules of P₂O₅

$\text{No. of molecules} = \text{0.0916 mol P _{2} O}_{5} \times \dfrac{6.022 \times 10^{23}\text{ molecules P _{2} O}_{5}}{\text{1 mol P _{2} O}_{5}}\\\\= \mathbf{5.5 \times 10^{22}}\textbf{ molecules P _{2} O}_{5}\\\text{There are \large \boxed{\mathbf{5.5 \times 10^{22}}\textbf{ molecules of P _{2} O}_{5}} }$