How would you balance: ${(N H_{4})}_{2} S O_{4} + K O H \to K_{2} S O_{4} + N H_{3} + H_{2} O$ $(NH_4)_2SO_4 + KOH -> K_2SO_4 + NH_3 + H_2O$ ?

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How would you balance: ${(N H_{4})}_{2} S O_{4} + K O H \to K_{2} S O_{4} + N H_{3} + H_{2} O$ $(NH_4)_2SO_4 + KOH -> K_2SO_4 + NH_3 + H_2O$ ?

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Answer 1 · 2016-01-02T15:03:37

You balance it stoichiometrically. What do I mean?

Explanation:

In every chemical reaction, both MASS and CHARGE are conserved. This is stoichiometry: for every product particle, there is a corresponding reactant particle.

So:

${(N H_{4})}_{2} S O_{4} (a q) + 2 K O H (a q) \to K_{2} S O_{4} (a q) + 2 N H_{3} (a q) + 2 H_{2} O (l)$ $(NH_4)_2SO_4(aq) + 2KOH(aq) rarr K_2SO_4(aq) + 2NH_3(aq) + 2H_2O(l)$

Does this reaction satisfy the condition of stoichiometry? Charge is balanced certainly, but what about mass? There are 10 $H$ $H$ particles as reactants. Are there 10 $H$ $H$ products? If not, the reaction is not stoichiometric.

Stoichiometry is not only practised by chemists. It is practised by shop-keepers, bankers, and salesman. For every credit there must be a corresponding debit. If there is not then someone, (YOU!), has made a mistake.

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How would you balance: ${(N H_{4})}_{2} S O_{4} + K O H \to K_{2} S O_{4} + N H_{3} + H_{2} O$ $(NH_4)_2SO_4 + KOH -> K_2SO_4 + NH_3 + H_2O$ ?

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How would you balance: (NH_4)_2SO_4 + KOH -> K_2SO_4 + NH_3 + H_2O(NH4)2SO4+KOH→K2SO4+NH3+H2O(NH_4)_2SO_4 + KOH -> K_2SO_4 + NH_3 + H_2O?

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How would you balance: ${(N H_{4})}_{2} S O_{4} + K O H \to K_{2} S O_{4} + N H_{3} + H_{2} O$ $(NH_4)_2SO_4 + KOH -> K_2SO_4 + NH_3 + H_2O$ ?