If 16.00 g of O₂ reacts with 80.00 g NO, what is the excess reactant? *

A. O₂
B. NO

  B. NO

Explanation:

In each mole of NO, the weight of the oxygen is 16.00 g; the weight of the nitrogen is 14.01 g. Then in 80 g of NO, we have 42.65 g of oxygen and 37.35 g of nitrogen.

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In each mole of NO₂, the weight of the oxygen is 2(16.00) = 32 g. The weight of the nitrogen is 1(14.01) = 14.01 g.

From the available oxygen, we can produce ...

 (42.65 g +16 g)/(32.00 g/mol) = 1.83 mol of NO₂

From the available nitrogen, we can produce ...

 (37.35 g)/(14.01 g/mol) = 2.67 mol of NO₂

Clearly, the reaction is limited by the amount of available oxygen.

NO is the excess reactant.

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