Metal hydrides react with water to form hydrogen gas and the metal hydroxide. srh2(s) + 2 h2o(l) sr(oh)2(s) + 2 h2(g) you wish to calculate the mass of hydrogen gas that can be prepared from 5.00 g of srh2 and 5.47 g of h2o. (a) how many moles of h2 can be produced from the given mass of srh2?

The reaction forms 0.112 mol H_2.  

We have the masses of two reactants, so this is a limiting reactant problem.

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

Step 1. Gather all the information 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

Step 2. Calculate the moles of each reactant  

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

Step 3. Identify the limiting reactant

Calculate the moles of H_2 we can obtain from each reactant.  

From SrH_2: Moles of H_2 = 0.055 77 mol SrH_2 × (2 mol H_2 /1 mol SrH_2) = 0.112 mol H_2

From H_2O: Moles of H_2 = 0.3036 mol H_2O × (2 mol H_2/2 mol H_2O)

= 0.3036 mol H_2

SrH_2 is the limiting reactant because it gives the smaller amount of H_2.

It produces 0.112 mol H_2.