Question #c41c6

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Question #c41c6

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Answer 1 · 2016-08-14T23:48:18

210 g of sodium is required.

Explanation:

The equation for the reaction is

$2Na + H_{2} {SO}_{4} \to H_{2} + {Na}_{2} {SO}_{4}$ $"2Na" + "H"_2"SO"_4 → "H"_2 + "Na"_2"SO"_4$

98 g of sulfuric acid (1 mol) reacts with 46 g (2 mol) of $Na$ $"Na"$ .

6.0 M $H_{2} {SO}_{4}$ $"H"_2"SO"_4$ implies 6.0 × 98 g of the acid in a litre of solution.

Thus 750 mL of solution contains $0,750 L \times 6.0 mol/L \times 98 g = {441 g H}_{2} {SO}_{4}$ $"0,750 L × 6.0 mol/L × 98 g" = "441 g H"_2"SO"_4$ .

This 441 g of $H_{2} {SO}_{4}$ $"H"_2"SO"_4$ reacts with ${441 g H}_{2} {SO}_{4} \times \frac{46 g Na}{{98 g H}_{2} {SO}_{4}} = 210 g Na$ $"441 g H"_2"SO"_4 × "46 g Na"/("98 g H"_2"SO"_4) = "210 g Na"$

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