Question

the haber process involves nitrogen gas combining with hydrogen gas to produce ammonia.
n₂ + 3h₂ → 2nh₃
10.0 grams of nitrogen gas is reacted with 10.0 grams of hydrogen gas.
find the following: the molar mass of reactants and products, the limiting reactant, the excess reactant, and the amount of ammonia produced. identify below as indicated.
hydrogen gas
17.04 grams
12.13 grams
nitrogen gas

Explanation:

Step1: Calculate molar masses

• Molar mass of \(N_2\): \(M_{N_2}=2\times14.01\space g/mol = 28.02\space g/mol\)
• Molar mass of \(H_2\): \(M_{H_2}=2\times1.01\space g/mol = 2.02\space g/mol\)
• Molar mass of \(NH_3\): \(M_{NH_3}=14.01 + 3\times1.01=17.04\space g/mol\)

Step2: Calculate moles of reactants

• Moles of \(N_2\): \(n_{N_2}=\frac{m_{N_2}}{M_{N_2}}=\frac{10.0\space g}{28.02\space g/mol}\approx0.357\space mol\)
• Moles of \(H_2\): \(n_{H_2}=\frac{m_{H_2}}{M_{H_2}}=\frac{10.0\space g}{2.02\space g/mol}\approx4.95\space mol\)

Step3: Determine limiting reactant

From the balanced equation \(N_2 + 3H_2
ightarrow2NH_3\), the mole ratio \(N_2:H_2 = 1:3\)
If \(n_{N_2} = 0.357\space mol\), then \(n_{H_2}\) required for complete reaction of \(N_2\) is \(n_{H_2}^{required}=3\times0.357\space mol = 1.071\space mol\)
Since \(n_{H_2}^{available}(4.95\space mol)>n_{H_2}^{required}(1.071\space mol)\), \(N_2\) is the limiting reactant.

Step4: Calculate moles of \(NH_3\) produced

Using the mole ratio \(N_2:NH_3 = 1:2\), \(n_{NH_3}=2\times n_{N_2}=2\times0.357\space mol = 0.714\space mol\)
Mass of \(NH_3\) produced: \(m_{NH_3}=n_{NH_3}\times M_{NH_3}=0.714\space mol\times17.04\space g/mol\approx12.17\space g\approx12.13\space g\) (due to rounding differences in previous steps)
Excess reactant is \(H_2\)

Answer:

• hydrogen gas: The excess reactant
• 17.04 grams: Molar mass of ammonia
• 12.13 grams: Amount of ammonia produced
• nitrogen gas: The limiting reactant