`One way to make ammonia is to synthesize it directly from elemental nitrogen and hydrogen (though this isn't that easy). The equation for this reaction would be N2 + 3 H2 → 2 NH3. If you are able to stream in 7.0 g of N2, what would be the minimum amount of H2 in grams that would be required to completely react with this amount of N2?

A. 1.5 g
B. 0.5 g
C. 0.75 g
D. 3.0 g
E. none of the above

Respuesta :

Answer:  A. 1.5 g

Explanation:

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

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

[tex]\text{Number of moles of nitrogen}=\frac{7.0g}{28g/mol}=0.25 moles[/tex]

According to stoichiometry:

1 mole of [tex]N_2[/tex] requires = 3 moles of [tex]H_2[/tex]

Thus 0.25 moles of [tex]N_2[/tex] will require =[tex]\frac{3}{1}\times 0.25=0.75moles[/tex] of [tex]H_2[/tex]

Mass of [tex]H_2[/tex] required =[tex]moles\times {\text {Molar mass}}=0.75mol\times 2g/mol=1.5g[/tex]

The minimum amount of [tex]H_2[/tex] in grams that would be required to completely react with this amount of [tex]N_2[/tex] is 1.5 grams.

Answer:

The correct answer is option A.

Explanation:

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

Moles of nitrogen gas = [tex]\frac{7.0 g}{28 g/mol}=0.25 mol[/tex]

According to reaction, 1 mole of nitrogen reacts with 3 moles of hydrogen gas.

Then 0.25 moles of nitrogen gas will react with:

[tex]\frac{3}{1}\times 0.25 mol=0.75 mol[/tex] of hydrogen gas.

Mass of 0.75 moles of hydrogen gas = 0.75 mol × 2 g/mol = 1.5 g

1.5 grams of hydrogen that would be required to completely react with this amount of nitrogen.

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