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Title: What mass of barium chloride is needed to react completely with 37.6 grams of sodium phosphate? Post by: smooth15 on Feb 15, 2013 The balanced equation is 3BaCl2 + 2Na3PO4 --> 6NaCl + Ba3(PO4)2. I really need some help, thanks in advance!
Title: What mass of barium chloride is needed to react completely with 37.6 grams of sodium phosphate? Post by: Ilee0405 on Feb 15, 2013 Since the molar mass of sodium phosphate is 3(23) + 94.97 + 4(16) = 227.97 g/mol, we see that require:
(37.6 g Na3PO4) * (1 mol Na3PO4)/(227.97 g Na3PO4) = 0.027 mol Na3PO4. From the balanced equation, we need: (0.027 mol Na3PO4) * (3 mol BaCl2)/(2 mol Na3PO4) = 0.0405 mol BaCl2. Then, since the molar mass of barium chloride is 13.734 + 2(35.45) = 84.63 g/mol, we require: (0.0405 mol BaCl2) * (84.63 g BaCl2)/(1 mol BaCl2) = 3.42 g BaCl2. I hope this helps! |