Question #35208

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?

Expert's answer

You can calculate this by such reaction:

NH_{₄}NO_{₃} --> N_{₂} + ½O_{₂} + 2H_{₂}O ==> 2NH_{₄}NO_{₃} --> 2N_{₂} + O_{₂} + 4H_{₂}O

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 N_{2} and 8 grams of O_{2}, then add them together to get 22 grams. Hence

m(N_{2}) = 14 g;

m(O_{2}) = 8 g;

In the sum it will be

m(N_{2} + O_{2}) = 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(H_{2}O) = m(NH_{₄}NO_{₃}) - m(N_{2} + O_{2}) = 40 - 22 = 18 grams

Answer: 18 grams of water will be formed

NH

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 N

m(N

m(O

In the sum it will be

m(N

Subtract the 22 from 40 grams and that gives you 18 grams left which is the water. In general it will be:

m(H

Answer: 18 grams of water will be formed

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