How many mL of 0.20 M CaCl2 are needed to react with 15 mL of 0.32 M K3PO4, give

Question

How many mL of 0.20 M CaCl2 are needed to react with 15 mL of 0.32 M K3PO4, given the balanced equation:
2 K3PO4 (aq) + 3 CaCl2 (aq) --> 6 KCl (aq) + Ca3(PO4)2 (s) How many mL of 0.20 M CaCl2 are needed to react with 15 mL of 0.32 M K3PO4, given the balanced equation:
2 K3PO4 (aq) + 3 CaCl2 (aq) --> 6 KCl (aq) + Ca3(PO4)2 (s) How many mL of 0.20 M CaCl2 are needed to react with 15 mL of 0.32 M K3PO4, given the balanced equation:
2 K3PO4 (aq) + 3 CaCl2 (aq) --> 6 KCl (aq) + Ca3(PO4)2 (s)

Explanation / Answer

from equation :

2 mol K3PO4 = 3 mol CaCl2

n1 = 2      ,   n2 = 3

M1V1/n1 = M2V2/n2

(0.32*15/2) = (0.2*V2/3)

V2 = volume of 0.20 M CaCl2 are needed = 36 ml

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