saldivarmia25
contestada

A chemist mixes 24.0 g H2 with 8.0 g N2. Assuming the reaction goes to completion, which reactant should she use to calculate the yield?
3H2 + N2 + 2NH3
A. Either reactant, because each will be completely used up in the reaction
B. Either reactant, because they are in a ratio of 3:1 for H2:N2
C. Hydrogen, because it could produce the greater amount of product
D. Nitrogen, because the ratio of moles is greater than 3:1 for Hz:N2​

Respuesta :

sebassandin

Answer:

D. Nitrogen, because the ratio of moles is greater than 3:1 for H2:N2

Explanation:

Hello,

In this case, given the reaction:

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

In order to identify the limiting reactant we must compute the moles of ammonia yielded by both reactants at first:

[tex]n_{NH_3}^{by\ H_2}=24.0gH_2*\frac{1molH_2}{2molH_2}*\frac{2molNH_3}{3molH_2}=8molNH_3\\ \\n_{NH_3}^{by\ N_2}=8.0gN_2*\frac{1molN_2}{28molN_2}*\frac{2molNH_3}{1molN_2}=0.57molNH_3[/tex]

Thus, since nitrogen yields the smallest amount of ammonia as it is more heavy than hydrogen and it is in a 3:1 mole ratio for H2:N2, it is the limiting reactant, therefore the answer is D. Nitrogen, because the ratio of moles is greater than 3:1 for H2:N2.

Regards.

Otras preguntas