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Consider the reaction below. How many grams of Ca3(PO4)2 will be produced if 7.89 g of Na3PO4 react completely? 3 Ca(NO3)2 (aq) + 2 Na3PO4 (aq) –> Ca3(PO4)2 (s) + 6 NaNO3 (aq) Na3PO4 = 163.94 g/mol Ca3(PO4)2 = 310.18 g/mol
Expert Answer
Answer
First, get mol of Na3PO4 so we can relate furhter amounts
mol = mass/MW = 7.89 / (163.94) = 0.048127 mol of Na3PO4
from the balanced reaction
2 mol of Na3PO4 = 1 mol of Ca3(PO4)2
then
0.048127 mol of Na3PO4 = 1/2*0.048127 = 0.024063 mol of Ca3(PO4)2
now,
change to mass
1 mol of Ca3(PO4)2 = 310.18 g
0.024063 mol of Ca3(PO4)2 = 310.18*0.024063 = 7.463 g of Ca3(PO4)2
