Answer to Question #35208 in Chemistry for lia

Question

When 40 grams of ammonium nitrate explode 14 grams of nitrogen and 8 grams of oxygen are formed. How many grams of water are formed?

Answer 1

You can calculate this by such reaction:

NH_₄NO_₃ --> N_₂ + ½O_₂ + 2H_₂O ==> 2NH_₄NO_₃ --> 2N_₂ + O_₂ + 4H_₂O

The law of conservation of mass solves this problem. If you start off with 40 grams you have to end up with 40 grams in a chemical reaction. If you produce 14 grams of N₂ and 8 grams of O₂, then add them together to get 22 grams. Hence
m(N₂) = 14 g;
m(O₂) = 8 g;
In the sum it will be
m(N₂ + O₂) = 14 + 8 = 22 grams
Subtract the 22 from 40 grams and that gives you 18 grams left which is the water. In general it will be:
m(H₂O) = m(NH_₄NO_₃) - m(N₂ + O₂) = 40 - 22 = 18 grams

Answer: 18 grams of water will be formed