Ishwar S. answered • 09/25/18
Tutor
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University Professor - General and Organic Chemistry
Hello Radi
The balanced chemical reaction is:
3Ca(OH)2 + 2H3PO4 → Ca3(PO4)2 + 6H2O
a) Use stoichiometry for this calculation. Mole ratio of Ca(OH)2 to Ca3(PO4)2 is 3:1. Molar mass of Ca(OH)2 = 74.09 g/mol and Ca3(PO4)2 = 310.18 g/mol.
25.0 g Ca(OH)2 x (1 mol Ca(OH)2 / 74.09 g Ca(OH)2) x (1 mol Ca3(PO4)2 / 3 mol Ca(OH)2) x (310.18 g Ca3(PO4)2 / 1 mol Ca3(PO4)2) = 34.9 g Ca3(PO4)2
b) First determine whether Ca(OH)2 or H3PO4 is the limiting reagent. Convert 0.139 moles of Ca(OH)2 to moles of H3PO4. If the calculated moles of H3PO4 is less than 0.58, then Ca(OH)2 is the limiting reagent whereas if it is more than 0.58, then H3PO4 is the limiting reagent. (NOTE! You can also perform a similar calculation starting with 0.58 moles of H3PO4 instead of Ca(OH)2 to determine the limiting reagent).
0.139 mol Ca(OH)2 x (2 mol H3PO4 / 3 mol Ca(OH)2) = 0.0927 mol H3PO4
You can see that only 0.0927 moles of H3PO4 are needed to react completely with 0.139 moles of Ca(OH)2. As a result, Ca(OH)2 is the limiting reagent. Grams of H2O produced is
0.139 mol Ca(OH)2 x (6 mol H2O / 3 mol Ca(OH)2) x (18.02 g H2O / 1 mol H2O) = 5.01 g H2O
Moles of excess H3PO4 remaining = 0.58 - 0.0927 = 0.49 moles (rounded to 2 decimal places after subtraction)
Grams of excess H3PO4 remaining = 0.49 mol H3PO4 x (97.99 g H3PO4 / 1 mol H3PO4) = 48 g H3PO4
