Be sure to answer all parts. Metal hydrides react with water to form hydrogen gas and metal hydroxide. For example, SrH2(s) + 2H2O(l) → Sr(OH)2(s) + 2H2(g) You wish to calculate the mass of hydrogen gas that can be prepared from 5.64 g of SrH2 and 4.70 g of H2O. (a) How many moles of H2 can be produced from the given mass of SrH2?

Explanation:

SrH₂ + 2H₂O = Sr(OH)₂ + 2H₂

90gm     36gm                  2 moles

5.64 g        4.7 g

water required for 5.64 g of SrH₂ = (36/ 90) x 5.64 g

= 2.256 g

water is in excess . Hence limiting reagent is SrH₂

90g SrH₂  makes 2 mole of water

5.64g SrH₂ makes water equal to mole = 2 x 5.64 / 90

= .125 mole .

mole of hydrogen formed = .125 .

the answer is d. 74.02