# Answer to Question #35208 in Other Chemistry for lia

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

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

NH

_{₄}NO_{₃}--> N_{₂}+ ½O_{₂}+ 2H_{₂}O ==> 2NH_{₄}NO_{₃}--> 2N_{₂}+ O_{₂}+ 4H_{₂}OThe 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. Hencem(N

_{2}) = 14 g;m(O

_{2}) = 8 g;In the sum it will be

m(N

_{2}+ O_{2}) = 14 + 8 = 22 gramsSubtract 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 gramsAnswer: 18 grams of water will be formed

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