posted by John on .
If 720 mL of 0.458 M aqueous Ba(OH)2 and 720 mL of 0.431 M aqueous HClO4 are reacted stoichiometrically according to the balanced equation, how many milliliters of 0.458 M aqueous Ba(OH)2 remain? Round your answer to 3 significant figures.
Ba(OH)2(aq) + 2HClO4(aq) → Ba(ClO4)2(aq) + 2H2O(l)
Make sure the equation is balanced.
moles Ba(OH)2 = M x L.
moles HClO4 = M x L.
From the question, I assume Ba(OH)2 is in excess and HClO4 is the limiting reagent. If so, use the coefficients in the balanced equation to convert moles HClO4 to moles Ba(OH)2 used. The difference between initial moles Ba(OH)2 and moles Ba(OH)2 used gives moles Ba(OH)2 remaining. Then M = moles/L. You know moles and M, calculate L remaining.