Review this reaction: $H_{2} S O_{4} + N a O H$ $H_2SO_4 + NaOH$ . What are th product(s)?

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Review this reaction: $H_{2} S O_{4} + N a O H$ $H_2SO_4 + NaOH$ . What are th product(s)?

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Answer 1 · 2017-04-05T23:39:28

The products are $H_{2} O$ $H_2O$ and $N a_{2} S O_{4}$ $Na_2SO_4$ , since this is a double-replacement/neutralization reaction.

Explanation:

This is a double-replacement neutralization (acid-base) reaction:

The two $H$ $H$ atoms in $H_{2} S O_{4}$ $H_2SO_4$ would react with the hydroxide ion, $O H^{-}$ $OH^-$ , in $N a O H$ $NaOH$ to form water, $H_{2} O$ $H_2O$ .

The $S O_{4}$ $SO_4$ would react with $N a$ $Na$ to form $N a_{2} S O_{4}$ $Na_2SO_4$ .

$H_{2} S O_{4} + N a O H \to H_{2} O + N a_{2} S O_{4}$ $H_2SO_4+NaOH->H_2O +Na_2SO_4$

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Review this reaction: $H_{2} S O_{4} + N a O H$ $H_2SO_4 + NaOH$ . What are th product(s)?

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Review this reaction: $H_{2} S O_{4} + N a O H$ $H_2SO_4 + NaOH$ . What are th product(s)?