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:
NH₄NO₃ --> N₂ + ½O₂ + 2H₂O ==> 2NH₄NO₃ --> 2N₂ + O₂ + 4H₂O
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(NH₄NO₃) - m(N2 + O2) = 40 - 22 = 18 grams
Answer: 18 grams of water will be formed
NH₄NO₃ --> N₂ + ½O₂ + 2H₂O ==> 2NH₄NO₃ --> 2N₂ + O₂ + 4H₂O
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(NH₄NO₃) - m(N2 + O2) = 40 - 22 = 18 grams
Answer: 18 grams of water will be formed
