The mass of ammonia in grams produced when 2. | vedclass

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(A) The balanced chemical equation for the reaction is: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$ Molar mass of $N_2 = 28 \ g/mol$,$H_2 = 2 \ g/mol$,$NH

Vedclass · Accepted Answer

$3400$

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(A) The balanced chemical equation for the reaction is: $N_2(g) + 3H_2(g) ightarrow 2NH_3(g)$ Molar mass of $N_2 = 28 \ g/mol$,$H_2 = 2 \ g/mol$,$NH_3 = 17 \ g/mol$. Given: $2.8 \ kg$ of $N_2 = 2800 \ g = 100 \ mol$. Given: $1 \ kg$ of $H_2 = 1000 \ g = 500 \ mol$. According to the stoichiometry,$1 \ mol$ of $N_2$ requires $3 \ mol$ of $H_2$. Therefore,$100 \ mol$ of $N_2$ requires $300 \ mol$ of $H_2$. Since $500 \ mol$ of $H_2$ is available,$N_2$ is the limiting reagent. $1 \ mol$ of $N_2$ produces $2 \ mol$ of $NH_3$. So,$100 \ mol$ of $N_2$ produces $200 \ mol$ of $NH_3$. Mass of $NH_3 = 200 \ mol 	imes 17 \ g/mol = 3400 \ g$.

The mass of ammonia in grams produced when $2.8 \ kg$ of dinitrogen quantitatively reacts with $1 \ kg$ of dihydrogen is.............

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