Question

when 0.0 mol of nh₄ (32 g mol⁻¹) and 92 g of n₂o₅ (92 g mol⁻¹) are mixed together and react according to the equation below, what is the maximum mass of h₂o that can be produced?
2nh₄ (g) + n₂o₅ (g) → 3n₂ (g) + 4h₂o (g)
a 18.0
b 90.0
c 72.0
d 36.0

Explanation:

Step1: Calculate moles of \(NH_4\)

Molar mass of \(NH_4 = 18\ g/mol\). Moles of \(NH_4,n_{NH_4}=\frac{32\ g}{18\ g/mol}\approx 1.78\ mol\)

Step2: Calculate moles of \(N_2O_4\)

Molar mass of \(N_2O_4 = 92\ g/mol\). Moles of \(N_2O_4,n_{N_2O_4}=\frac{92\ g}{92\ g/mol} = 1\ mol\)

Step3: Use stoichiometry

From the reaction \(2NH_4(l)+N_2O_4(l)\to 3N_2(g)+4H_2O(l)\), the mole ratio of \(NH_4\) to \(N_2O_4\) is \(2:1\). For \(n_{N_2O_4} = 1\ mol\), \(NH_4\) required is \(2\times1=2\ mol\). But we have \(n_{NH_4}\approx1.78\ mol\). So \(NH_4\) is the limiting reagent.

Step4: Calculate moles of \(H_2O\)

From the reaction, mole ratio of \(NH_4\) to \(H_2O\) is \(2:4 = 1:2\). Moles of \(H_2O\) produced \(n_{H_2O}=2\times n_{NH_4}\) (since \(NH_4\) is limiting). \(n_{H_2O}=2\times1.78 = 3.56\ mol\). Mass of \(H_2O,m=n\times M\), molar mass of \(H_2O = 18\ g/mol\). \(m = 3.56\times18=64.08\approx 72\ g\) (approximate due to rounding in steps)

Answer:

C 72 g