Answer:
[tex]\boxed{1.26 \times 10^{20} \text{ molecules NH}_{3}}[/tex]
Explanation:
We will need a balanced chemical equation with masses, moles, and molar masses.
1. Gather all the information in one place:
M_r: 2.016 17.03
3H₂ + N₂ ⟶ 2NH₃
m/g: 6.33 × 10⁻⁴
2. Calculate the moles of H₂
[tex]\text{Moles of H}_{2} = 6.33 \times 10^{-4}\text{ g H}_{2} \times \dfrac{\text{1 mol H}_{2}}{\text{2.016 g H}_{2}} = 3.140 \times 10^{-4} \text{ mol H}_{2}[/tex]
3. Calculate the moles of NH₃
The molar ratio is 2 mol NH₃/3 mol H₂.
[tex]\text{Moles of NH}_{3} = 3.140 \times 10^{-4} \text{ mol H}_{2} \times \dfrac{\text{2 mol NH}_{3}}{\text{3 mol H}_{2}} = 2.093 \times 10^{-4}\text{ mol NH}_{3}[/tex]
4. Calculate the molecules of NH₃
There are 6.022 × 10²³ molecules of NH₃/1 mol NH₃.
[tex]\text{Molecules of NH}_{3} = 2.093 \times 10^{-4}\text{ mol NH}_{3} \times \dfrac{6.022 \times 10^{23}\text{ molecules NH}_{3}}{\text{1 mol NH}_{3}}\\= 1.26 \times 10^{20}\text{ molecules NH}_{3}\\\text{The reaction produces } \boxed{\mathbf{1.26 \times 10^{20}} \textbf{ molecules NH}_{3}}[/tex]
