Hydrogen gas, H2, reacts with nitrogen gas, N2, to form ammonia gas, NH3, according to the equation

3H2(g)+N2(g)→2NH3(g)

NOTE: Throughout this tutorial use molar masses expressed to five significant figures.
Part A
How many moles of NH3 can be produced from 18.0 mol of H2 and excess N2?
Part B
How many grams of NH3 can be produced from 4.50 mol of N2 and excess H2.
Part C
How many grams of H2 are needed to produce 12.57 g of NH3?
Part D
How many molecules (not moles) of NH3 are produced from 1.99×10−4 g of H2?

A. 12.0 moles of ammonia can be produced.

B. 153.28 grams of ammonia can be produced.

C. 2.2318 grams of hydrogen gas will be needed.

D. molecules of ammonia will be produced.

Explanation:

A.

Moles of hydrogen gas = 18.0 mol

According to reaction, 3 moles of hydrogen gas gives 2 moles of ammonia:

Then 18.0 moles of hydrogen gas will give ;

of ammonia

12.0 moles of ammonia can be produced.

B.

Moles of nitrogen gas = 4.50 mol

According to reaction, 1 moles of nitrogen gas gives 2 moles of ammonia:

Then 4.50 moles of nitrogen gas will give ;

of ammonia

Molar mass of ammonia = 17.031 g/mol

Mass of 9.0 moles of ammonia:

9.0 mol × 17.031 g/mol = 153.28 g

153.28 grams of ammonia can be produced.

C.

Mass of ammonia gas = 12.57 g

Moles of ammonia gas =

According to reaction, 2 moles of ammonia are obtained from 3 moles of hydrogen gas.

Then 0.73806 moles of ammonia gas will be obtained from;

of hydrogen gas

Molar mass of hydrogen gas= 2.0159 g/mol

Mass of 1.1071 moles of hydrogen :

1.1071 mol ×  2.0159 g/mol = 2.2318 g

2.2318 grams of hydrogen gas will be needed.

D.

Mass of hydrogen gas =

Moles of hydrogen gas =

According to reaction, 3 moles of hydrogen gas gives 2 moles of ammonia:

Then moles of hydrogen gas will give ;

of ammonia.

1 mole = molecules/ atoms

Molecules of ammonia in .

molecules of ammonia will be produced.

the correct answer is:

a. formation of precipitation