In Haber's process 50.0 g of $N_{2}$ (g) and 10.0g of $H_{2}$ (g) are mixed to produce $N H_{3}$ (g). What the number of moles of $N H_{3}$ (g) formed?

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$2.36$

$5.36$

$2.01$

$3.33$

answer is A.

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Detailed Solution

Number of moles of nitrogen $= w e i g h t / m o l e c u l a r w e i g h t = 50 / 28 = 1.785$ moles

Number of moles of hydrogen $= w e i g h t / m o l e c u l a r w e i g h t = 10 / 2 = 5$ moles

The balanced equation is

$N_{2} + 3 H_{2} \to 2 N H_{3} 1 3 2 m o l e s g i v e n, 1.785 5 ?$

Here hydrogen is the limiting reagent. As it is known that three mole of

hydrogen to form two moles of ammonia.

Then five moles of hydrogen will produce $2 / 3 \times 5 = 3.33$

Hence the correct option is (A).

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