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

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asked by Sarah
  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.

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