Question

question 4 (1 point)
for the reaction 3 h₂ (g) + n₂ (g) → 2 nh₃ (g), 3 mol h₂ is reacted with 6 mol n₂,
how many grams of nh₃ are formed?
57 g
26 g
34 g
68 g
17 g

Explanation:

Step1: Determine the limiting reactant

The balanced equation is \(3H_2(g)+N_2(g)\to2NH_3(g)\). The mole ratio of \(H_2\) to \(N_2\) is \(3:1\). Given \(n(H_2) = 3mol\) and \(n(N_2)=6mol\). For \(3molH_2\), the required \(N_2\) is \(1mol\) (since \(3molH_2\times\frac{1molN_2}{3molH_2}\)). So \(H_2\) is the limiting reactant.

Step2: Calculate moles of \(NH_3\)

From the balanced equation, the mole ratio of \(H_2\) to \(NH_3\) is \(3:2\). So \(n(NH_3)=3molH_2\times\frac{2molNH_3}{3molH_2}=2mol\)

Step3: Calculate mass of \(NH_3\)

The molar mass of \(NH_3\) is \(M = 17g/mol\). Using \(m=n\times M\), \(m = 2mol\times17g/mol = 34g\)

Answer:

34 g