The first step in the ostwald process for producing nitric acid is as follows: 4nh3 (g) + 5o2 (g) ? 4no (g) + 6h2o (g). if the reaction of 15.0 g of ammonia with 15.0 g of oxygen gas yields 8.70 g of nitric oxide, what is the percent yield of this reaction?

Question

Mckinley Norman

Chemistry

18 July, 00:41

The first step in the ostwald process for producing nitric acid is as follows: 4nh3 (g) + 5o2 (g) ? 4no (g) + 6h2o (g). if the reaction of 15.0 g of ammonia with 15.0 g of oxygen gas yields 8.70 g of nitric oxide, what is the percent yield of this reaction?

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Higgins · Answer 1

4 moles of NH3 reacts with 5 moles of O2.

4 moles of NH3 = 4 (14 + 3*1) = 68g

5 moles of O2 = 5*32 = 160g

Therefore, 68g of NH3 reacts with 160g of O2

Mass of NH3 to react with 15g of oxygen = 15*68 / 160 = 6.375g, which is less than 15 g.

Mass of O2 to react with 15g of NH3 = 15*160 / 68 = 35.29g, which is more than 15g.

Therefore, the limiting reactant is O2, which means the reaction will proceed until the 15 g of oxygen finish.

4 moles of NO = 4 (14+16) 120g

Therefore, 160g of O2 gives 120g of NO.

15g of O2 will give Xg of NO.

X (theoritical yeild) = 15*120 / 160 = 11.25g

Practical yield = 8.70g

Yield = practical yeild / theoritical yeild * 100 = 8.70/11.25 = 77.33%