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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

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Kalahira
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

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