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Question


2NO(g)+O2(g)2NO2(g)+150 kJ
If 60 g of NO and 60 g of O2 are available, what is the maximum amount of NO2 that can be produced?
(Molecular weights: NO=30 g/mole, NO2=46 g/mole)

A
46 g
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B
60 g
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C
92 g
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D
120 g
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E
180 g
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Solution

The correct option is C 92 g
Mass is divided with molecular weight to obtain number of moles.
60 grams of NO =6030=2 moles.
60 grams of O2=6032=1.875 moles.
2 moles of NO will react with 1 mole of O2. However 1.875 moles of O2 are present. Hence, O2 is excess reagent and NO is limiting reagent.
2 moles of NO will produce 2 moles of NO2.
The molecular weight of NO2 is 46 g/mol.
Mass of NO2 produced is 2×46=92 grams.

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