Hydrogen cyanide gas can be made by a two stepprocess. First, ammonia is reacted with oxygen gas to give nitrogen monoxide and water vapor. In the next step, nitrogen monoxide is reacted with methane (CH4) to give hydrogen cyanide gas. The by-products are water and hydrogen gas. A) Write the balanced equation for the two reactions. B) When 24.2 g of ammonia and 25.1 g of methane are used, how many grams of hydrogen cyanide can be produced?

38.34 g of HCN are produced

Explanation:

Our reactants for the first reaction are:

NH₃ and O₂

Products are: H₂O  and NO

Our reactants for the second reaction are:

NO and CH₄

Producs are: H₂, H₂O and HCN

The reactions are:

4NH₃ and 5O₂  →  6H₂O  + 4NO

2NO and 2CH₄  →  2HCN  + H₂ + 2H₂O

In the first step 4 moles of ammonia can produce 4 moles of NO, so ratio is 1:1

24.2 g . 1mol / 17 g = 1.42 moles of ammonia

We have produced 1.42 moles of NO.

25.1 g . 1 mol / 16 g = 1.57 moles of methane.

Ratio is 2:2. So, for 1.57 moles of methane, we need 1.57 moles of NO.

Moles are the same. As there is no enough NO, this is the limiting reactant.

Ratio with product is also 2:2.

Our 1.42 moles of NO have produced 1.42 moles of HCN.

We convert moles to mass: 1.42 mol . 27 g/mol = 38.34 g

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explanation:

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