What volume of .0500 M Ba(OH)2 will react completely with 21.00 mL of .500 M HCl?

Question

Same type problem. Post your work if you get stuck.

2HCl + Ba(OH)2 --> 2HOH + BaCl2
(.0500 mol HCl / 1000 mL) (21.00 mL) (1 mol Ba(OH)2 / 2 mol HCl) (1000 mL / .0500 mol Ba(OH)2)

= 1.05e2 mL Ba(OH)2

right.