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of nitrogen gas)?
How many grams of H2 are needed to produce 71.1 g of ammonia (NH3) (assuming unlimited availability
3 H2+ N2 + 2NH3
Provide the answer with 3 or more significant figures.

Respuesta :

sebassandin

Answer:

[tex]m_{H_2}=12.6gH_2[/tex]

Explanation:

Hello there!

In this case, given the reaction:

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

It is possible to compute the necessary grams of hydrogen which produce 71.1 g of ammonia, given the 3:2 mole ratio and their molar mass as shown below:

[tex]m_{H_2}=71.1gNH_3*\frac{1molNH_3}{17.04gNH_3}*\frac{3molH_2}{2molNH_3} *\frac{2.02gH_2}{1molH_2}\\\\m_{H_2}=12.6gH_2[/tex]

Best regards!

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