Consider the balanced chemical equation: 3Sr(OH)2(aq) + 2H3PO4(aq) → Sr3(PO4)2(s) + 6H2O(l). How many moles of Sr(OH)2 remain if 0.10 mol of H3PO4 react with 0.40 mol of Sr(OH)2?

Answer 1

First you must determine the limiting reagent but I PRESUME that the limiting reagent must be H3PO4 (otherwise no Sr(OH)2 would be left). Then convert 0.1 mole H3PO4 to moles Sr(OH)2.
0.1 mole H3PO4 x (3 moles Sr(OH)2/2 moles H3PO4) = 0.1 x (3/2) = ?? moles H3PO4 used.
Then 0.4 - moles used = moles remaining.

Answer 2

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Answer 3

0.35

Answer 4

0.25

Answer 5

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