Question

how many grams of h₂ are needed to produce 13.56 g of nh₃? express your answer numerically in grams. view available hint(s) g h₂ submit previous answers incorrect; try again; 5 attempts remaining part d how many molecules (not moles) of nh₃ are produced from 2.77×10⁻⁴ g of h₂? express your answer numerically as the number of molecules. view available hint(s) molecules

Explanation:

Step1: Write the balanced chemical equation

$N_2 + 3H_2
ightarrow2NH_3$

Step2: Calculate the molar mass of $NH_3$ and $H_2$

The molar mass of $NH_3$ is $14 + 3\times1= 17$ g/mol, and the molar mass of $H_2$ is $2\times1 = 2$ g/mol.

Step3: Calculate moles of $NH_3$ in the first - part

The number of moles of $NH_3$, $n_{NH_3}=\frac{m_{NH_3}}{M_{NH_3}}=\frac{13.56}{17}=0.7976$ mol.

Step4: Determine moles of $H_2$ needed using stoichiometry

From the balanced equation, the mole - ratio of $H_2$ to $NH_3$ is 3:2. So, $n_{H_2}=\frac{3}{2}n_{NH_3}=\frac{3}{2}\times0.7976 = 1.1964$ mol.

Step5: Calculate mass of $H_2$

$m_{H_2}=n_{H_2}\times M_{H_2}=1.1964\times2 = 2.3928\approx2.39$ g.

Step6: For the second - part, calculate moles of $H_2$

$n_{H_2}=\frac{m_{H_2}}{M_{H_2}}=\frac{2.77\times 10^{-4}}{2}=1.385\times10^{-4}$ mol.

Step7: Determine moles of $NH_3$ using stoichiometry

From the balanced equation, $n_{NH_3}=\frac{2}{3}n_{H_2}=\frac{2}{3}\times1.385\times 10^{-4}=9.233\times10^{-5}$ mol.

Step8: Calculate the number of molecules of $NH_3$

The number of molecules of $NH_3$, $N = n_{NH_3}\times N_A$, where $N_A = 6.022\times10^{23}$ molecules/mol. So, $N=9.233\times10^{-5}\times6.022\times10^{23}=5.56\times10^{19}$ molecules.

Answer:

2.39 g
5.56×10^{19} molecules