Question #1935e

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Question #1935e

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Answer 1 · 2015-01-12T01:04:54

The balanced equation is:

$H_{2} S O_{4 (a q)} + 2 N H_{4} O H_{(a q)} \to {(N H_{4})}_{2} S O_{4} (a q) + 2 H_{2} O_{(l)}$ $H_2SO_(4(aq))+2NH_4OH_((aq))rarr(NH_4)_2SO_4(aq)+2H_2O_((l))$

Sulfuric acid is diprotic so is able to give up 2 H+ ions. $N H_{4} O H$ $NH_4OH$ is not a compound that can actually exist as a discrete substance. It is better described as ammonia solution which is formed from dissolving ammonia gas in water:

$N H_{3 (g)} + H_{2} O_{(l)} ⇌ N H_{4 (a q)}^{+} + O H_{(a q)}^{-}$ $NH_(3(g))+ H_2O_((l))rightleftharpoonsNH_(4(aq))^(+)+OH_((aq))^-$

Answer 2 · 2015-01-12T01:07:35

You are dealing with a reaction between a strong acid and a weak base, which results in the formation of salt and water. The reaction's general equation is

$H_{2} S O_{4 (a q)} + 2 N H_{4} O H_{(a q)} \to {(N H_{4})}_{2} S O_{4} (a q) + 2 H_{2} O_{(l)}$ $H_2SO_(4(aq)) + 2NH_4OH_((aq)) -> (NH_4)_2SO_4(aq) + 2H_2O_((l))$

The complete ionic equation looks like this

$2 H_{(a q)}^{+} + S O_{4 (a q)}^{2 -} + 2 N H_{4 (a q)}^{+} + 2 O H_{(a q)}^{-} \to 2 N H_{4 (a q)}^{+} + S O_{4 (a q)}^{2 -} + 2 H_{2} O_{(l)}$ $2H_((aq))^(+)+ SO_(4(aq))^(2-) + 2NH_(4(aq))^(+) + 2OH_((aq))^(-) -> 2NH_(4(aq))^(+) + SO_(4(aq))^(2-) + 2H_2O_((l))$

The net ionic equation you'll often see is

$H_{(a q)}^{+} + O H_{(a q)}^{-} \to H_{2} O_{(l)}$ $H_((aq))^(+) + OH_((aq))^(-) -> H_2O_((l))$

However, sulfuric acid, which is a strong acid, reacts with ammonium hydroxide, which is a weak base, to produce salt, ammonium sulfate, and water.

Since it is a weak base, ammonium hydroxide will dissociate only slightly in water, so another version of the net ionic equation could be:

$H_{(a q)}^{+} + N H_{4} O H_{(a q)} \to N H_{4 (a q)}^{+} + H_{2} O_{(l)}$ $H_((aq))^(+) + NH_4OH_((aq)) -> NH_(4(aq))^(+) + H_2O_((l))$

However, this does not change the balanced general reaction, which still stands as it was written in the beginning of the post.

Answer 3 · 2015-01-12T01:10:07

$H_{2} S O_{4} + 2 N H_{4} O H \to {(N H_{4})}_{2} S O_{4} + 2 H_{2} O$ $H_2SO_4+2NH_4OH->(NH_4)_2SO_4+2H_2O$

You see that at the right of the equation there are 2 $N H_{4}$ $NH_4$ -groups, while at the left there is only one.

To even this out you may only double the whole $N H_{4} O H$ $NH_4OH$
which works out right, because then the 2 $H$ $H$ 's from the $H_{2} S O_{4}$ $H_2SO_4$ can combine with the 2 $O H$ $OH$ 's from the $N H_{4} O H$ $NH_4OH$
to form 2 $H_{2} O$ $H_2O$ 's

In reality it all happens a bit differently, with ions and stuff like that, but for balancing the equation this is enough.

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Question #1935e

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