Using the balanced equation, perform the stoichiometry calculations indicated in a. and b. below 2 H3PO4 + 3 Sr -> 3 H2 + Sr3(PO4)2

a. How many grams of strontium phosphate are produced in the reaction of 65.30 g of strontium with excess phosphoric acid?

b. How many grams of hydrogen are produced?

Thanks.

a) 113g Sr3(PO4)2

b) 1.5g H2

Explanation:

The number of moles of Sr used in the equation

= mass/ molar mass

=65.30/87

= 0.751mol

Since the mole ratio of Sr: Sr3(PO4)2 is 3:1

moles of Sr3(PO4)3 = 1/3*0.751

= 0.25mol

grams of Sr3(PO4)3 is = moles * molar mass

= 0.25 * 452

= 113g

b). Since mole ratio of Sr:H2 is 3:3

moles of H2 = 0.751 * 3/3 ---> 0.751 mol

grams of H2 is = moles * molar mass

= 0.751 * 2

= 1.502g

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