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 #"HCN"# that we can get from each reactant.
The balanced chemical equation is
#"2NH"_3 + "3O"_2 + "2CH"_4→ "2HCN" + "6H"_2"O"#
From #"NH"_3#:
#"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 #"O"_2#: #"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 #"CH"_4#:
#"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"#
#"O"_2# gives the fewest moles of #"HCN"#, so #"O"_2# is the limiting reactant.
#"Mass of HCN" = 0.250 cancel("mol HCN") × "27.03 g HCN"/(1 cancel("mol HCN")) = "6.76 g HCN"#
