Some Basic Concepts Of Chemistry Question 17
Question: $ BrO _3^{\odot }+5B{r^{\odot }}\to Br _2+3H _2O $ If 50 mL 0.1 M $ BrO _3^{\odot } $ is mixed with 30 mL of 0.5 M $ B{r^{\Theta }} $ solution that contains excess of $ {H^{\oplus }} $ ions, the moles of $ Br _2 $ formed are
Options:
A) $ 6.0\times {10^{-4}} $
B) $ 1.2\times {10^{-4}} $
C) $ 9.0\times {10^{-3}} $
D) $ 1.8\times {10^{-3}} $
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Answer:
Correct Answer: C
Solution:
Balance the equation: $ \underset{1mmol}{\mathop{BrO _3^{\odot }}},+\underset{5mmol}{\mathop{5B{r^{\odot }}}},\to \underset{1mmol}{\mathop{3Br _2}},+3H _2O $ Given: $ 50\times 0.1 $ $ 30\times 0.5 $ =5 mmol =15 mmol $ B{r^{\odot }} $ is the limiting reagent. 5 mmol of $ B{r^{\odot }} $ gives
$ \Rightarrow $ 3 mmol of $ Br _2 $ 15 mmol of $ B{r^{\odot }} $ gives
$ \Rightarrow $ $ \frac{3\times 15}{5} $ =9 mmol = $ 9\times {10^{-3}}mol $