posted by josselyne .
How many milliliters of 0.762 M H3PO4 are required to react with 243 mL of 0.159 M Ba(OH)2 if the products are barium phosphate and water?
3Ba(OH)2 + 2H3PO4 ==> Ba3(PO4)2 + 6H2O
mols Ba(OH)2 = M x L = ?
Convert mols Ba(OH)2 to mols H3PO4 using the coefficients in the balanced equation.
The M H3PO4 = mols H3PO4/L H3PO4. YOu know mols H3PO4 and M H3PO4, solve for L H3PO4 and convert to mL.