`BrO_(3)^(ɵ) + 5Br^(ɵ) rarr Br_(2) + 3 H_(2) O`
IF `50 mL 0.1 M BrO_(3)^(ɵ)` is mixed with `30 mL` of `0.5 M Br^(ɵ)` solution that contains excess of `H^(o+)` ions, the moles of `Br_(2)` formed are

Options

A. `6.0 xx 10^(4)`
B. `1.2 xx 10^(-4)`
C. `9.0 xx 10^(-3)`
D. `1.8 xx 10^(-3)`

Answer

C

Explanation

Balance of equation:
Given : `underset({:(1 "mmol"),(50 xx 0.1),(= 5 "mmol"):})(BrCO_(3)^(ɵ)) + underset({:(5 "mmol"),(30 xx 0.5),(= 15 "mmol"):})(5Br^(ɵ)) rarr underset(1 "mmol")(3Br_(2)) + 3H_(2) O`
`Br^(ɵ)` is the limiting reagent.
5 mmol of `Br^(ɵ)` gives `implies` 3 mmol of `Br_(2)`
15 mmol of `Br^(ɵ)` gives `implies (3 xx 15)/(5) = 9 "mmo"`
`9 xx 10^(-3) "mol"`