Question

How many grams of hydrogen are required to produce 68 grams of ammonia if 56 grams of nitrogen are taken initially?​

Answer 1

Explanation:

refer to the attachment

Answer 2

Given:

68 grams of ammonia is produced by 56 grams of nitrogen.

To find:

We have to find the amount of hydrogen.

Solution:

The reaction between nitrogen and hydrogen to form ammonia is given as-

$N_2+3H_2\rightarrow 2NH_3$.

Thus one mole of nitrogen reacts with 3 moles of hydrogen to form 2 moles of ammonia.

The molecular weight of nitrogen gas is 28 grams.

So, the number of moles occupy by 56 grams of nitrogen is 56/28=2 moles.

The molecular weight of ammonia is 17 grams.

Thus the number of moles occupied by 68 grams of ammonia is 68/17=4 moles.

So, one mole of nitrogen reacts with 3 moles of hydrogen to form 2 moles of ammonia.

So, 2 moles of nitrogen will react with 3×2=6 moles of hydrogen to form 4 moles of ammonia.

Thus, the mass of 1 mole of hydrogen gas is 2 grams.

So, the mass of 6 moles of hydrogen gas is 2×6=12 grams.

Thus 12 grams of hydrogen is used.