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NISA [10]
3 years ago
8

The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In t

he first step of the Ostwald process, ammonia is reacted with oxygen gas to produce nitric oxide and water. What is the maximum mass of H 2 O H2O that can be produced by combining 62.8 g 62.8 g of each reactant? 4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )
Chemistry
1 answer:
ICE Princess25 [194]3 years ago
3 0

Answer:

42,3g of H₂O

Explanation:

For the reaction:

4NH₃(g) + 5O₂(g) ⟶ 4NO(g) + 6H₂O(g)

62,8 g of NH₃ are:

62,8g×(1mol/17,031g) = <em>3,69 moles of NH₃</em>

62,8 g of O₂ are:

62,8g×(1mol/32g) =<em> 1,96 moles of O₂</em>

For a complete reaction of 1,96 moles of O₂ you need:

1,96mol O₂×(4mol NH₃ / 5molO₂) = 1,57 moles NH₃. As you have 3,69 moles, limiting reactant is O₂.

Assuming a complete reaction, 1,96mol O₂ produce:

1,96mol O₂×(6mol H₂O / 5molO₂) = 2,35 moles of H₂O. In grams:

2,35 moles of H₂O×(18,01g/1mol) = <em>42,3g of H₂O</em>

I hope it helps!

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