Answered

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?

Answer :

Answer:

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. [tex]3.963\times 10^{19} [/tex] molecules of ammonia will be produced.

Explanation:

[tex]3H_2(g)+N_2(g)\rightarrow 2NH_3(g)[/tex]

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 ;

[tex]\frac{2}{3}\times 18.0 mol=12.0 mol[/tex] 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 ;

[tex]\frac{2}{1}\times 4.50 mol=9.0 mol[/tex] 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 = [tex]\frac{12.57 g}{17.031 g/mol}=0.73806 mol[/tex]

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;

[tex]\frac{3}{2}\times 0.73806 mol=1.1071 mol[/tex] 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 = [tex]1.99\times 10^{-4} g[/tex]

Moles of hydrogen gas = [tex]\frac{1.99\times 10^{-4} g}{2.0159 g/mol}=9.871\times 10^{-5} mol[/tex]

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

Then [tex]9.871\times 10^{-5} mol[/tex]moles of hydrogen gas will give ;

[tex]\frac{2}{3}\times 9.871\times 10^{-5} mol=6.581\times 10^{-5} mol[/tex] of ammonia.

1 mole = [tex]N_A=6.022\times 10^{23}[/tex] molecules/ atoms

Molecules of ammonia in [tex]6.581\times 10^{-5} mol[/tex].

[tex]=6.581\times 10^{-5}\times 6.022\times 10^{23} molecules[/tex]

[tex]=3.963\times 10^{19} molecules[/tex]

[tex]3.963\times 10^{19} [/tex] molecules of ammonia will be produced.

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