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)

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.

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