Question #c41c6

1 Answer
Aug 14, 2016

Answer:

210 g of sodium is required.

Explanation:

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"#