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Angelina_Jolie [31]
3 years ago
8

According to the following reaction, how many grams of ammonia will be formed upon the complete reaction of 31.2 grams of hydrog

en gas with excess nitrogen gas ? nitrogen(g) + hydrogen(g) ammonia(g)
Chemistry
1 answer:
Aneli [31]3 years ago
8 0

Answer:

176.8 g of ammonia, NH3.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

N2 + 3H2 —> 2NH3

Next, we shall determine the mass of H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:

Molar mass of H2 = 2x1 = 2 g/mol

Mass of H2 from the balanced equation = 3 x 2 = 6 g

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 2 x 17 = 34 g.

From the balanced equation above,

6 g of H2 reacted to produce 34 g of NH3.

Finally, we shall determine the mass of ammonia, NH3 produced by reacting 31.2 g of H2.

This can be obtained as follow:

From the balanced equation above,

6 g of H2 reacted to produce 34 g of NH3.

Therefore, 31.2 g of H2 will react to produce = (31.2 x 34)/6 = 176.8 g of NH3.

Therefore, 176.8 g of ammonia, NH3 were obtained from the reaction.

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The change in internal energy for the combustion of 1.0 mol of octane at a pressure of 1.0 atm is -5084.1 kj . f the change in e
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<u>Given:</u>

Change in internal energy = ΔU = -5084.1 kJ

Change in enthalpy = ΔH = -5074.3 kJ

<u>To determine:</u>

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Therefore:

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W = ΔU-ΔH = -5084.1 -(-5074.3) = -9.8 kJ

Ans: Work done is -9.8 kJ


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