Question

part d
how many molecules (not moles) of nh₃ are produced from 8.06×10⁻⁴ g of h₂?
express your answer numerically as the number of molecules.
view available hint(s)

Explanation:

Step1: Calculate moles of H₂

The molar mass of H₂ is 2.016 g/mol. Use the formula $n=\frac{m}{M}$, where $n$ is the number of moles, $m$ is the mass, and $M$ is the molar mass. So, $n_{H_2}=\frac{8.06\times 10^{-4}\text{ g}}{2.016\text{ g/mol}}$.
$n_{H_2}=\frac{8.06\times 10^{-4}}{2.016}\text{ mol}\approx 4.00\times 10^{-4}\text{ mol}$

Step2: Use the stoichiometry

The balanced chemical equation for the production of NH₃ from H₂ and N₂ is $N_2 + 3H_2
ightarrow 2NH_3$. The mole - ratio of H₂ to NH₃ is 3:2. So, $n_{NH_3}=\frac{2}{3}n_{H_2}$.
$n_{NH_3}=\frac{2}{3}\times4.00\times 10^{-4}\text{ mol}\approx 2.67\times 10^{-4}\text{ mol}$

Step3: Calculate the number of molecules

Use Avogadro's number $N = n\times N_A$, where $N$ is the number of molecules, $n$ is the number of moles, and $N_A = 6.022\times 10^{23}\text{ molecules/mol}$.
$N_{NH_3}=2.67\times 10^{-4}\text{ mol}\times6.022\times 10^{23}\text{ molecules/mol}$
$N_{NH_3}=2.67\times6.022\times 10^{-4 + 23}\text{ molecules}\approx 1.61\times 10^{20}\text{ molecules}$

Answer:

$1.61\times 10^{20}$