One way to remove nitrogen oxide from smokestack emissions is to react it with ammonia: 4 nh3 (g) + 6 no (g) > 5 n2 (g) + 6h2o (i)
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12.3 mol of no reacts with mol of ammonia
5.87 mol of no yields mol nitrogen

The answer 12,3 mol of NO reacts with 8.2_mol of ammonia 5.87 mol of NO yields 4.89_ mol nitrogen Proof This kind of question is called mole-mole problem, for solving it we process as follow 4 NH3 (g) + 6 NO (g) > 5 N2 (g) + 6H2O 4moles 6moles X 12.3 moles This implies x=12.3 x 4 /6=8.2 moles, so 12.3 moles of nitrogen corresponds to 8.2 ammoniac 4 NH3 (g) + 6 NO (g) ---> 5 N2 (g) + 6H2O 6 moles 5moles . 5.87 moles x=? After cross-multiplication, 5.87 x 5 / 6 = 4.89, so 5.87 moles of nitrogen oxyde yelds 4.89 moles of nitrogen

B. a graduated substance