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Stoichiometric Calculations

Consider the ...

Question

Consider the following reaction:
$3 M g + N_{2} ⟶ M g_{3} N_{2}$ .
Calculate amount of $M g$ required to prepare $10$ g of $M g_{3} N_{2}$ in $100 %$ reaction.(If the answer is $x$ , then write $x \times 100$ ).

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Solution

The molar masses of $M g$ and $M g_{3} N_{2}$ are $24.30$ g/mol and $100.9$ g/mol respectively.
$3$ moles of $M g$ gives $1$ mole of $M g_{3} N_{2}$ .
Hence, $3 \times 24.30$ g of $M g$ will give $100.9$ g of $M g_{3} N_{2}$ .
Hence, the mass of $M g$ required to form $10$ g of $M g_{3} N_{2}$ is $10 \times \frac{3 \times 24.30}{100.9} = 7.2$ g.

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