You visited us 6 times! Enjoying our articles? Unlock Full Access!

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]

Open in App

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 $10 x = 2$ .

Suggest Corrections

Similar questions

3 M g + N_{2} ⟶ M g_{3} N_{2}

M g

10

M g_{3} N_{2}

100 %

x

x \times 100

M g_{3} N_{2}

(N = 14, M g = 24.)

M g_{3} N_{2}

M g_{3} N_{2}

N H_{3}

2 M g + O_{2} \to 2 M g O

3 M g + N_{2} \to M g_{3} N_{2}

M g_{3} N_{2} + 6 H_{2} O \to 3 M g (O H)_{2} + 2 N H_{3} ↑

M g_{3} N_{2}

48 g

M g

34 g m

N H_{3}

M g + N H_{3} \to M g_{3} N_{2} + H_{2}

N_{2} + 3 H_{2} \to 2 N H_{3}

4 N H_{3} + 5 O_{2} \to 4 N O + 6 H_{2} O

N_{2}

H_{2}

O_{2}

Join BYJU'S Learning Program