Question #c41c6

Aug 14, 2016

210 g of sodium is required.

Explanation:

The equation for the reaction is

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

98 g of sulfuric acid (1 mol) reacts with 46 g (2 mol) of $\text{Na}$.

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

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

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