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How many grams of nitrogen (N2) would be required to create 9.3 moles of ammonia (NH 3) in the following reaction: N2 +3H 2 →2NH 3?

A) 34.8 grams
B) 94.2 grams
C) 280.8 grams
D) 168.6 grams

User Colinmarc
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1 Answer

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Final answer:

To create 9.3 moles of ammonia, 34.8 grams of nitrogen (N2) would be required.

Step-by-step explanation:

To determine the number of grams of nitrogen (N2) required to produce 9.3 moles of ammonia (NH3), we need to use the mole ratio from the balanced equation:

N2(g) + 3H2(g) -> 2NH3(g)

From the equation, we can see that for every 1 mole of N2, 2 moles of NH3 are produced. Therefore, the mole ratio is 1:2.

Using the mole ratio, we can set up a proportion:

(9.3 mol NH3 / 2 mol NH3) = (x g N2 / 1 mol N2)

Solving for x, we find that x = 34.8 grams. Therefore, the correct answer is A) 34.8 grams.

User Karthik Kumar
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