If $19 \cdot m o l$ $19*mol$ of $H_{2} (g)$ $H_2(g)$ react with stoichiometric $N_{2} (g)$ $N_2(g)$ , what molar quantity of ammonia results?

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If $19 \cdot m o l$ $19*mol$ of $H_{2} (g)$ $H_2(g)$ react with stoichiometric $N_{2} (g)$ $N_2(g)$ , what molar quantity of ammonia results?

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Answer 1 · 2017-07-06T06:02:57

A bit under $13 \cdot m o l$ $13*mol$ of ammonia..........

Explanation:

You have the stoichiometric equation........

$N_{2} (g) + 3 H_{2} (g) catalysis - --- \to 2 N H_{3} (g)$ $N_2(g) + 3H_2(g) stackrel("catalysis")rarr2NH_3(g)$

And clearly, if $19 \cdot m o l$ $19*mol$ dihydrogen react completely (which is NOT an altogether probable outcome) then by the stoichiometry we get $\frac{19}{3} \times 2 \cdot m o l = 12.7 \cdot m o l$ $19/3xx2*mol=12.7*mol$ with respect to ammonia, $N H_{3}$ $NH_3$ .......

What is the significance of this ammonia synthesis? Do you think that this reaction is widely performed?

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If $19 \cdot m o l$ $19*mol$ of $H_{2} (g)$ $H_2(g)$ react with stoichiometric $N_{2} (g)$ $N_2(g)$ , what molar quantity of ammonia results?

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Explanation:

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If 19*mol19⋅mol19*mol of H_2(g)H2(g)H_2(g) react with stoichiometric N_2(g)N2(g)N_2(g), what molar quantity of ammonia results?

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Explanation:

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If $19 \cdot m o l$ $19*mol$ of $H_{2} (g)$ $H_2(g)$ react with stoichiometric $N_{2} (g)$ $N_2(g)$ , what molar quantity of ammonia results?