The mass of Mg_3N_2 produced if 48 g of Mg m | vedclass

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(A) The balanced chemical equation is: $3Mg + 2NH_3 	o Mg_3N_2 + 3H_2$ Stoichiometry calculations: $3 \ mol \ Mg = 3 	imes 24 \ g = 72 \ g$ $2 \ mol

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$\frac{200}{3} \ g$

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(A) The balanced chemical equation is: $3Mg + 2NH_3 	o Mg_3N_2 + 3H_2$ Stoichiometry calculations: $3 \ mol \ Mg = 3 	imes 24 \ g = 72 \ g$ $2 \ mol \ NH_3 = 2 	imes 17 \ g = 34 \ g$ $1 \ mol \ Mg_3N_2 = (3 	imes 24) + (2 	imes 14) = 100 \ g$ Since $48 \ g$ of $Mg$ is present and $34 \ g$ of $NH_3$ is present,$Mg$ acts as the Limiting Reagent $(L.R.)$. $72 \ g \ Mg$ produces $100 \ g \ Mg_3N_2$ $48 \ g \ Mg$ produces $\frac{100}{72} 	imes 48 \ g \ Mg_3N_2 = \frac{200}{3} \ g$

The mass of $Mg_3N_2$ produced if $48 \ g$ of $Mg$ metal is reacted with $34 \ g$ $NH_3$ gas is:
$3Mg + 2NH_3 \to Mg_3N_2 + 3H_2$

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The mass of $Mg_3N_2$ produced if $48 \ g$ of $Mg$ metal is reacted with $34 \ g$ $NH_3$ gas is: $3Mg + 2NH_3 \to Mg_3N_2 + 3H_2$