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Standard XII

Chemistry

Molarity

56 Kg of N2...

Question

56 Kg of $N_{2}$ gas and 10 Kg of $H_{2}$ gas are mixed to produce $N H_{3}$ gas. Calculate the no of moles of ammonia gas formed.

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Solution

$56$ kg of $N_{2}$
no. of moles of $N_{2}$ gas $= \frac{56 \times 10^{3}}{28} = 2 \times 10^{3}$ moles
$10$ kg of $H_{2}$
no. of moles of $H_{2}$ gas $= \frac{10 \times 10^{3}}{2} = 4.96 \times 10^{3}$ moles
$N_{2} + 3 H_{2} ⇌ 2 N H_{2}$
lesser moles for Hydrogen $\Rightarrow$ Hydrogen is limiting reagent
$3$ moles $H_{2} \to 2$ moles ${N H}_{3}$
$4.96$ moles $H_{2} \to \frac{(4.96) (2)}{3}$
$3.30$ moles of ${N H}_{3}$ are formed.

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What is the limiting reagent? $50 k g N_{2}$ and $10 k g H_{2}$ are mixed to produce $N H_{3}$ . Calculate the ammonia gas formed. Also, identify the limiting reagent.

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56 Kg of N2 gas and 10 Kg of H2 gas are mixed to produce NH3 gas. Calculate the no of moles of ammonia gas formed.

56kg of N2no. of moles of N2 gas =56×10328=2×103 moles10kg of H2no. of moles of H2 gas =10×1032=4.96×103 molesN2+3H2⇌2NH2lesser moles for Hydrogen ⇒ Hydrogen is limiting reagent3 moles H2→2 moles NH34.96 moles H2→(4.96)(2)33.30 moles of NH3 are formed.

56 Kg of $N_{2}$ gas and 10 Kg of $H_{2}$ gas are mixed to produce $N H_{3}$ gas. Calculate the no of moles of ammonia gas formed.