Question

How many milliliters of sodium metal, with a density of 0.97 g/mL, would be needed to produce 53.2 grams of hydrogen gas in the single-replacement reaction below? Show all steps of your calculation as well as the final answer.
Na + H2O → NaOH + H2

My answer:
1. Balance the equation: 2Na + 2H2O --> 2NaOH + H2
2. Take the known amount of 54.2 g H2 and convert it into moles, multiply by the mole ratio (number of moles from the balanced equation, unknown substance over the given substance) and convert it back into grams:
53.2 g H2 x 1 mol H2/ 2 g H2 x 2 moles Na/ 1 mol H2 x 22.98 g/ 1 mol Na = 1222 g Na
1 gram = 1 mL,
so 1222 mL of Na

Answer 1

The balanced chemical reaction is:

2Na + 2H2O → 2NaOH + H2

We first use the amount of hydrogen gas to be produced and the molar mass of the hydrogen gas to determine the amount in moles to be produced. Then, we use the relation from the reaction to relate H2 to Na.

53.2 g H2 ( 1 mol / 2.02 g ) ( 2 mol Na / 1 mol H2 ) ( 22.99 g / 1 mol ) = 1210.96 g Na

1210.96 g Na ( 1 mL / 0.97 g ) = 1248.41 mL Na needed