Asked by Anonymous
How many milliliters of 0.736 M H3PO4 are required to react with 238 mL of 0.160 M Ba(OH)2 if the products are barium phosphate and water?
Answers
Answered by
DrBob222
2H3PO4 + 3Ba(OH)2 ==> Ba3(PO4)2 + 6H2O
mols Ba(OH)2 = M x L = ?
Use the coefficients in the balanced equation to convert mols Ba(OH)2 to mols H3PO4.
That is ?mols Ba(OH)2 x [2 mol H3PO4/3 mol Ba3(PO4)2] = xx
Now M H3PO4 = mols H3PO4/L H3PO4.l You know M and mols, solve for L and convert to mL.
mols Ba(OH)2 = M x L = ?
Use the coefficients in the balanced equation to convert mols Ba(OH)2 to mols H3PO4.
That is ?mols Ba(OH)2 x [2 mol H3PO4/3 mol Ba3(PO4)2] = xx
Now M H3PO4 = mols H3PO4/L H3PO4.l You know M and mols, solve for L and convert to mL.
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