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How many mL of 0.40 M barium chloride are needed to react 175 mL of 1.50 M ammonium phosphate in the precipitation of barium phosphate?

Sagot :

OverpoweredNoob

Answer:

656.25 mL

Explanation:

M1 = 0.40 M

M2 = 1.50 M

V2 = 175 mL

Plug those values into the equation:

M1V1 = M2V2

(0.40)V1 = (1.50)(175)

---> V1 = 656.25 mL

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