Question #5641d

1 Answer
Ernest Z.
Aug 24, 2015

The solution contains 8.17 g of "H"_3"PO"_4.

Explanation:

"Equiv. mass" = "molar mass"/"no of ionizable H" = "98.00 g"/3 = "32.67 g"

"Mass of H"_3"PO"_4 = 0.500 color(red)(cancel(color(black)("L"))) × (0.500 color(red)(cancel(color(black)("eq"))))/(1 color(red)(cancel(color(black)("L")))) × "32.67 g"/(1 color(red)(cancel(color(black)("eq")))) = "8.17 g"

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