Question

For $4{NH}_3\left(g\right)+5O_2\left(g\right)\to 4NO\left(g\right)+6H_{2}O\left(g\right)$, if you begin with 66.00 g ammonia and 54.00 g oxygen, how many grams of water will be obtained?

• A. 2.294
• B. 36.51
• C. 1.409
• D. 25.3
• E. 2.513

Answer 1

Answer: (B)

36.51

The molecular weights of ammonia, oxygen and water are 17 g/mole, 32 g/mole and 18 g/mole respectively.
66.00 g ammonia $=\frac{66.00}{17}=3.88$ moles ammonia.
54.00 g oxygen $=\frac{54.00}{32.0}=1.6875$ moles oxygen.
4 moles ammonia reacts with 5 moles oxygen.
Hence, 3.88 moles ammonia will react with $3.88\times \frac{5}{4}=4.85$ moles oxygen but only 1.6875 moles oxygen are present. Hence oxygen is the limiting reagent.
5 moles of oxygen gives 6 moles of water.
1.728 moles of oxygen will give $1.6875\times \frac{6}{5}=2.025$ moles of water.
The mass of water obtained is $18\times 2.025=36.51$ g of water.