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Answer 1 · 2016-08-14T23:48:18

210 g of sodium is required.

The equation for the reaction is

#"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"#.

6.0 M #"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 × 6.0 mol/L × 98 g" = "441 g H"_2"SO"_4#.

This 441 g of #"H"_2"SO"_4# reacts with #"441 g H"_2"SO"_4 × "46 g Na"/("98 g H"_2"SO"_4) = "210 g Na"#