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The maximum amount of hydrogen gas that can be prepared is if all the hydrogen from both compounds is released.

The hydrogen in 4.94 g of SrH2 is calculated from the mass ratios between Sr and H

1) H2 in SrH2

Sr atomic mass = 87.62 g/mol

H2 molar mass = 2.02 g/mol

Mass of 1 mol of SrH2 = 87.62 g / mol + 2.02 g/mol = 89.64 g/mol

Ratio of H2 to SrH2 = 2.02 g H2 / 89.64 g SrH2

Proportion: 2.02 g H2 / 89.64 gSrH2 = x / 4.93 g SrH2

=> x = 4.93 g SrH2 * 2.02 g H2 / 89.64 g SrH2 = 0.111 g H2

2) H2 in H2O

2.02 g H2 / 18.02 g H2O * 4.14 g H2O = 0.464 g H2

3) Total mass of hydrogen = 0.111 g + 0.464 g = 0.575 g

Answer: 0.575 g

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