Question #d56f3

Apr 18, 2015

The maximum mass of hydrogen cyanide that could form is 6.76 g.

This is a limiting reactant problem, so we calculate the amount of $\text{HCN}$ that we can get from each reactant.

The balanced chemical equation is

$\text{2NH"_3 + "3O"_2 + "2CH"_4→ "2HCN" + "6H"_2"O}$

From ${\text{NH}}_{3}$:

$\text{Moles of HCN" = 11.5 cancel("g NH₃") × (1 cancel("mol NH₃"))/(17.03 cancel("g NH₃")) × "2 mol HCN"/(2 cancel("mol NH₃")) = "0.675 mol HCN}$

From ${\text{O}}_{2}$: $\text{Moles of HCN" = 12.0 cancel("g O₂") × (1 cancel("mol O₂"))/(32.00 cancel("g O₂")) × "2 mol HCN"/(3 cancel("mol O₂")) = "0.250 mol HCN}$

From ${\text{CH}}_{4}$:

$\text{Moles of HCN" = 10.5 cancel("g CH₄") × (1 cancel("mol CH₄"))/(16.04 cancel("g CH₄")) × "2 mol HCN"/(2 cancel("mol CH₄")) = "0.655 mol HCN}$

${\text{O}}_{2}$ gives the fewest moles of $\text{HCN}$, so ${\text{O}}_{2}$ is the limiting reactant.

$\text{Mass of HCN" = 0.250 cancel("mol HCN") × "27.03 g HCN"/(1 cancel("mol HCN")) = "6.76 g HCN}$