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
