A) \[6.0\times {{10}^{-4}}\]
B) \[1.2\times {{10}^{-4}}\]
C) \[9.0\times {{10}^{-3}}\]
D) \[1.8\times {{10}^{-3}}\]
Correct Answer: C
Solution :
Balance the equation: \[\underset{1\,\,mmol}{\mathop{Br{{O}_{3}}^{\odot }}}\,+\underset{5\,\,mmol}{\mathop{5B{{r}^{\odot }}}}\,\to \underset{1\,\,mmol}{\mathop{3B{{r}_{2}}}}\,+3{{H}_{2}}O\] 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 \[B{{r}_{2}}\] 15 mmol of \[B{{r}^{\odot }}\]gives\[\Rightarrow \]\[\frac{3\times 15}{5}\]=9 mmol =\[9\times {{10}^{-3}}mol\]
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