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Stoichiometric Calculations

Consider the ...

Question

Consider the following reaction:

$3 M g + N_{2} ⟶ M g_{3} N_{2}$ ;

If the number of moles of $M g_{3} N_{2}$ formed from $5.6$ g of $N_{2}$ with excess of $M g$ in $100 %$ reaction is $x$ . The value of $10 x$ is :
[Given that the molecular weight of $M g = 24, N = 14$ g/mol]

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Solution

The molar mass of nitrogen is $28$ g/mol.
The number of moles present in $5.6$ g of nitrogen is $\frac{5.6 g}{28 g / m o l} = 0.2 m o l$ .
$1$ mole of nitrogen forms $1$ mole of $M g_{3} N_{2}$ .
Hence, $0.2$ moles of nitrogen will form $0.2$ moles of $M g_{3} N_{2}$ .
Hence, $x = 0.2$ .
Therefore, the value of $10 x = 2$ .

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