Nitric oxide, NO, is made from the oxidation of NH_3, and the reaction it repres

Question

Nitric oxide, NO, is made from the oxidation of NH_3, and the reaction it represented by the equation: 4NH_3 + 5O_2 righatrrow 4NO + 6H_2O A chemist used 16.4 g of O_2 in this reaction and got 9.0 g of purified NO. the percent yield of NO is 78% 63% 41% 73% 20% SO_2 reacts with H_2S as follows: 2H_2S + SO_2 rightarrow 3S + 2H_2O When 7.5 g of H_2S reacts with 12.8 g of SO_2. which statement applies? 19.2 g of sulfur are formed. 10.6 g of sulfur are formed. 7.1 g of sulfur are formed. 6.4 g of sulfur are formed. None of the above are correct.

Explanation / Answer

12)

4NH3 + 5O2 --> 4NO + 6H2O

As per the equation 4 moles of NH3 will react with 5 moles of O2 to give 4 moles of NO

Or we can say that 160 grams of oxygen will react with 68 g of NH3 to give 120 grams of NO

So given oxygen amount will react to give = 120 X 16.4 / 160 grams of NO = 12.3 grams of NO (theoretically)

Actual amount obtained = 9 grams

So % yield = Actual amount X 100 / Theoretical amount = 73.17 % (option c)

13) reaction is

2H2S + SO2 --> 3S + 2H2O

So as per stoichiometry

2 moles of H2S should react with 1 mole of SO2 to give three moles of sulphur

Molecular weight of H2S = 34g /mole

Molecular weight of SO2 = 64 g / mole

atomic weight weight of S = 32 g / mole

So the weight ratio of H2S : SO2 should be equal to = 2X 34 / 64 = 1.06

Given ratio of H2S : SO2 = 7.5 / 12.8 = 0.59

so the amount of H2S is less as compared to SO2 hence it will be the limiting reagent

We can say that

68 grams of H2S reacts with 64 grams of SO2 to give 96 grams of Sulphur

Or 1 gram of H2S will react with 64/68 grams of SO2 to give 96 / 68 grams of sulphur

Or 7.5 grams of H2S will react with 64 X 7.5 / 68 grams of SO2 to give 96 X 7.5 / 68 grams of sulphur

7.5 grams of H2S will react with 7.06 grams of SO2 to give 10.59 grams of Sulphur

hence correct statement is 10.6 grams of sulphur will be obtained

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