When solutions of sodium phosfate and barium nitrate are mixed, a precipitate of barium
phospfate is formed according to the equation
2 Na3PO4(aq) + 3 Ba(NO3)2(aq) ---> Ba3(PO4)2(s) + 6 NaNO3(aq)
What mass of Ba3(PO4)2 is formed when a solution containing 410 g of Na3PO4 is mixed with a solution containing an excess of Ba(NO3)2?
The Attempt at a Solution
is it 1505.1g ?