Ammonia contains 82.65 % $N_{2}$ and 17.65% $H_{2}$ . If the law of constant proportions is true, then the mass of zinc required to give 10 g ammonia will be:

8.265 g

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0.826 g

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82.65 g

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826.5 g

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Solution

The correct option is D 0.826 g
$N_{2} + H_{2} \to 2 N H_{3}$
Every $17$ gms of $N H_{3}$ will contain $14$ gms of Nitrogen and $3$ gms of Hydrogen
(Mol. man of ${N H}_{3}$ = $14 + 3 (1) = 17$ )
moles of $N_{2}$ = $\frac{5}{17}$
Max of $N_{2}$ = $\frac{528}{17}$
= $8.235 g$ of nitrogen

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Ammonia contains 82.65 % N2 and 17.65% H2. If the law of constant proportions is true, then the mass of zinc required to give 10 g ammonia will be:

The correct option is D 0.826 gN2+H2→2NH3Every 17 gms of NH3 will contain 14 gms of Nitrogen and 3 gms of Hydrogen (Mol. man of NH3= 14+3(1)=17)moles of N2 = 517Max of N2= 52817=8.235g of nitrogen

Ammonia contains 82.65 % $N_{2}$ and 17.65% $H_{2}$ . If the law of constant proportions is true, then the mass of zinc required to give 10 g ammonia will be:

The correct option is D 0.826 g
$N_{2} + H_{2} \to 2 N H_{3}$
Every $17$ gms of $N H_{3}$ will contain $14$ gms of Nitrogen and $3$ gms of Hydrogen
(Mol. man of ${N H}_{3}$ = $14 + 3 (1) = 17$ )
moles of $N_{2}$ = $\frac{5}{17}$
Max of $N_{2}$ = $\frac{528}{17}$
= $8.235 g$ of nitrogen