If 5.00 grams of each ammonia, oxygen and methane (ch4) gas react, how much hydrogen cyanide can be produced

if 2.10 grams of HCN is produced what is the percent yield? water is the other product

First write out the reaction:

NH3(g) + O2(g) + CH4 (g) ----> HCN + H2O

Balance the equation accordingly..

4 NH3(g) + 6 O2(g) + 4 CH4 (g) ----> 4 HCN + 12 H2O

Given a mass of each, you have to find which one will be the limiting reactant. First convert to moles

5/(16.0* 2 ) [remember oxygen is diatomic] = .156 moles O2

5/(3* 1.0079 + 14.007) = .239 moles NH3

5/(12.011 + 4 * 1.0079) = .312 moles CH4

Looking at the chemical equation, it can be concluded that O2 will be your limiting reactant, since you need 6 moles per reaction vs. 4 for the other two compounds.

Using a proportion:

6 mole O2/4 mol HCN =.156 mol O2/x mol HCN

Cross multipyling and solving for x gives

6x = .156 * 4
x = .104 mols HCN should be formed according to the math

Converting to grams gives:
.104 mols HCN * (1.0079+12.011+14.007) = 2.81 grams should be formed

However, only 2.10 grams were.. so:

2.10g/2.81g x 100 = 74.7 % yield.


The biggest trick to the problem is getting the chemical equation right. The rest is stoichiometry.