What is the mass of hydrogen produced when 8.0 g of sodium reacts with an excess water to produce hydrogen and sodium hydroxide? 2 Na + 2 H2O -> 2NaOH + H2 (Please help, thank you!)

Answer 1

Since we have 2 reactants, the question already states that water is the excess reagent. Therefore, the amount of H₂ 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 H₂ using the mole ratio from the balanced reaction. From the reaction, you can see that it takes 2 mol Na to form 1 mol H₂.

0.35 mol Na x 1 mol H₂ / 2 mol Na = 0.18 mol H₂

Finally, convert mol of H₂ to g of H₂ using its molar mass (2 g/mol)

0.18 mol H₂ x 2 g H₂ / 1 mol H₂ = 0.36 g H₂