If 200 mL of 3.70 M aqueous Ba(OH)2 and 980 mL of 0.837 M aqueous HClO4 are reacted stoichiometrically according to the balanced equation, how many milliliters of 3.70 M aqueous Ba(OH)2 remain? Round your answer to 3 significant figures.

Question

Ba(OH)2(aq) + 2HClO4(aq) → Ba(ClO4)2(aq) + 2H2O(l)

answer = 89.2....

my calculations:
Ba(OH)2 = 0.74 su
HClO4 = 0.410 su
so... HClO4 is the limiting reagent

what do i do next?....

Answer 1

Good work so far.
So you have 0.740 moles Ba(OH)2
You have 0.410 moles HClO4 reacting
What's left is 0.740-0.410 = 0.330 moles Ba(OH)2.
M = moles/L
You know M and moles, solve for L and convert to mL, then round to three s.f.