you wish to calculate the mass of hydrogen gas that can be prepared from 4.89 g of SrH2 and 5.28 g of H2O.

Question

(a) How many moles of H2 can be produced from the given mass of SrH2?

Answer 1

This is a limiting reagent problem. I know that because an amount is given for both reactants.
SrH2 + 2H2O ==> Sr(OH)2 + 2H2
mols SrH2 = 4.89/molar mass SrH2.
mols H2O = 5.28/molar mass H2O.

Using the coefficients in the balanced equation, convert mols SrH2 to mols H2.
Do the same for mols H2O to mols H2.

It is likely the two values will not be the same which means one of them is not right. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.

Now convert mols H2 to grams. g = mols x molar mass.