The reaction forms 0.112 mol H_2.
We have the masses of two reactants, so this is a <em>limiting reactant problem</em>.
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.
<em>Step 1. Gather all the information</em> in one place with molar masses above the formulas and everything else below them.
M_r: __89.64 ___18.02 ___________2.016
______SrH_2 + 2H_2O → Sr(OH)_2 + 2H_2
Mass/g: 5.00 ___5.47
<em>Step 2</em>. Calculate the <em>moles of each reactant </em>
Moles of SrH_2 = 5.00 g SrH_2 × (1 mol SrH_2 /89.64 g SrH_2)
= 0.055 77 mol SrH_2
Moles of H_2O = 5.47 g H_2O × (1 mol H_2O/18.02 g H_2O)
= 0.3036 mol H_2O
<em>Step 3</em>. Identify the <em>limiting reactant </em>
Calculate the moles of H_2 we can obtain from each reactant.
<em>From SrH_2</em>: Moles of H_2 = 0.055 77 mol SrH_2 × (2 mol H_2 /1 mol SrH_2) = 0.112 mol H_2
<em>From H_2O</em>: Moles of H_2 = 0.3036 mol H_2O × (2 mol H_2/2 mol H_2O)
= 0.3036 mol H_2
<em>SrH_2 is the limiting reactant</em> because it gives the smaller amount of H_2.
It produces 0.112 mol H_2.