Solve the following limiting reactant problem: Ca 3 (PO 4 ) 2 +3H 2 SO 4 3CaSO 4

Question

Solve the following limiting reactant problem: Ca3(PO4)2 +3H2SO4 3CaSO4 +2H3PO4 17.3g              15.5g a.What is the limiting reactant? b. what is the excess reactant? c. how many grams of calcium sulfate are formed in thisreaction? Solve the following limiting reactant problem: Ca3(PO4)2 +3H2SO4 3CaSO4 +2H3PO4 17.3g              15.5g a.What is the limiting reactant? b. what is the excess reactant? c. how many grams of calcium sulfate are formed in thisreaction? 17.3g              15.5g a.What is the limiting reactant? b. what is the excess reactant? c. how many grams of calcium sulfate are formed in thisreaction?

Explanation / Answer

If you have 17.3 g Calcium Phosphate, that is 17.3 g x 1 mol/310 g= .0558 mol If you have 15.5 g sulfuric acid, that is 15.5 g x 1 mol/98g = .158mol Since you need 3 mols of sulfuric acid for every mole of calciumphosphate, divide .158 by 3 and see if it is greater than.0558. It is not. It is smaller. That meanssulfuric acid is the limiting reactant. That means Calciumphosphate is the excess reactant. To find how many moles of calcium sulfate are formed, do thefollowing: .158 mol sulfuric acid x 3 mol calcium sulfate/3 mol sulfuric acidx 136 g/1 mol calcium sulfate = 21.5 g

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