Question

How many milliliters of manganese metal, with a density of 7.43 g/mL, would be needed to produce 21.7 grams of hydrogen gas in the single-replacement reaction below? Show all steps of your calculation as well as the final answer. Mn + H2O → MnO + H2 (5 points)

Answer 1

First compute the number of grams of manganese metal required to make 21.7 grams of H2.
21.7 g H2 x (1 mole H2/ 2 g H2) x (1 mole Mn/1 mol H2) x (55 grams Mn/1 mol Mn) = 596.75 grams
Now density = mass/volume
7.43 = 596.75/volume
volume = 596.75/7.43 = 80.31 mL
80.31 mL is the amount of manganese needed.