Since we have 2 reactants, the question already states that water is the excess reagent. Therefore, the amount of H_{2} gas formed is dependent on the mass of sodium (Na; 8.0 g) that you start with.

Since the reaction is already balanced, convert mass of Na to moles of Na using its molar mass (23 g/mol).

8.0 g x (1 mol Na / 23 g Na) = 0.35 mol Na (rounded to 2 significant figures)

Next, convert mol Na to mol H_{2} using the mole ratio from the balanced reaction. From the reaction, you can see that it takes 2 mol Na to form 1 mol H_{2}.

0.35 mol Na x 1 mol H_{2} / 2 mol Na = 0.18 mol H_{2}

Finally, convert mol of H_{2} to g of H_{2} using its molar mass (2 g/mol)

0.18 mol H_{2} x 2 g H_{2} / 1 mol H_{2} = **0.36 g H**_{2}