How many grams of $H_{2}$ $H_2$ must react, if the reaction needs to produced 63.5g of $N H_{3}$ $NH_3$ ?

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How many grams of $H_{2}$ $H_2$ must react, if the reaction needs to produced 63.5g of $N H_{3}$ $NH_3$ ?

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Answer 1 · 2017-02-06T17:39:42

We need approx. $12 \cdot g$ $12*g$ of $dihydrogen gas.........$ $"dihydrogen gas........."$

Explanation:

As a preliminary we require a stoichiometrically balanced equation:

$\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) ⇌ N H_{3} (g)$ $1/2N_2(g) + 3/2H_2(g) rightleftharpoonsNH_3(g)$

Is this balanced?

$Moles of ammonia = \frac{63.5 \cdot g}{17.01 \cdot g \cdot m o l^{- 1}} = 3.73 \cdot m o l .$ $"Moles of ammonia"=(63.5*g)/(17.01*g*mol^-1)=3.73*mol.$

Given the stoichiometric equation, clearly, we need $\frac{3}{2}$ $3/2$ $equiv$ $"equiv"$ of $dihydrogen gas:$ $"dihydrogen gas:"$

$i.e.$ $"i.e."$ $\frac{3}{2} \times 3.73 \cdot m o l \times 2.02 \cdot g \cdot m o l^{- 1} = 11.3 \cdot g$ $3/2xx3.73*molxx2.02*g*mol^-1=11.3*g$

How much dinitrogen do we need?

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How many grams of $H_{2}$ $H_2$ must react, if the reaction needs to produced 63.5g of $N H_{3}$ $NH_3$ ?

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Explanation:

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How many grams of $H_{2}$ $H_2$ must react, if the reaction needs to produced 63.5g of $N H_{3}$ $NH_3$ ?