Balance the reaction below in acidic conditions: IO3- + I- --> I2

Question

What volume of 0.352 M HCl is needed to produce 2.48x10-3 moles of iodine, I2, with an excess of KlO3 and Kl?

Answer 1

IO3^- + 5I^- + 6H^+ ==> 3I2 + 3H2O

moles I2 needed = 0.00248.
moles HCl needed to produce 0.00248 moles I2. 0.00248 x (6 moles HCl/3moles I2) = 0.00248 x (6/3) = ??
Them M x L = moles. YOu know moles and M, solve for L.