Question #814fe

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Question #814fe

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Answer 1 · 2016-05-23T06:01:00

This question is not well-proposed, in that sodium metal will react with the water solvent as well.

Explanation:

But for sodium hydroxide, $N a O H$ $NaOH$ , we can write the followig reaction:

$2 N a O H (a q) + H_{2} S O_{4} (a q) \to N a_{2} S O_{4} (a q) + 2 H_{2} O (a q)$ $2NaOH(aq) + H_2SO_4(aq) rarr Na_2SO_4(aq) +2H_2O(aq)$

$Moles of sulfuric acid$ $"Moles of sulfuric acid"$ $=$ $=$ $0.750 \cdot L \times 6.0 \cdot m o l \cdot L^{- 1}$ $0.750*Lxx6.0*mol*L^-1$ $=$ $=$ $4.50 \cdot m o l$ $4.50*mol$ .

Thus we need to add $9.0 \cdot m o l$ $9.0*mol$ sodium hydroxide to neutralize the sulfuric acid. $9.0 \cdot m o l \times 40.0 \cdot g \cdot m o l^{- 1}$ $9.0*molxx40.0*g*mol^-1$ $=$ $=$ $360 \cdot g$ $360*g$ .

Why do we need to add $9$ $9$ $m o l$ $mol$ caustic soda rather than $4.5$ $4.5$ $m o l$ $mol$ ?

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Question #814fe

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