This is a limiting reagent problem.
1. Write the equation and balance it.
SrH2 + 2H2O ==> Sr(OH)2 + 2H2 Check me out on that.
2a. Convert 3.78 g SrH2 to moles. moles = grams/molar mass.
2b. Convert 3.5 g H2O to moles.
3a. Using the balanced equation, convert moles SrH2 to moles H2.
3b. Using the balanced equation, convert moles H2O to moles H2.
3c. You have two answers for the same product and one of them MUST be wrong. The correct one, as in ALL limiting reagent problems, is the smaller one. The reactant producing that smaller number is the limiting reagent.
4. The moles from 3c x molar mass = grams produced.
metal hydrides react with water to form hydrogen gas and the metal hydroxide. How do you calculate the mass of hydrogen gas that can be prepared from 3.78 g of SrH2 and 3.5 g of H2O?
a) how many moles of H2 can be produced from given mass of SrH2?
b) How many grams of H2?
THANK YOU!!
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