Question

how many moles of KBrO3 are required to prepare 0.0700 moles of Br2 according to the reaction :
KBrO3+ 5KBr+ 6HNO3➡️6KNO3 + 3Br2+3H2O
(a) 0.210
(b) 0.0732
(c) 0.0704
(d) 0.0233​

Answer 1

Answer:

the answer is c

0.0704

correct answer

Answer 2

Given:

Reaction $\[KBr{O_3} + 5KBr + 6HN{O_3} \to 6KN{O_3} + 3B{r_2} + 3{H_2}O\]$

Number of moles of $\[B{r_2}\]$

To Find: Number of moles of $\[KBr{O_3}\]$ are required to prepare 0.0700 moles of $\[B{r_2}\]$

Solution:

According to the given reaction $\[KBr{O_3} + 5KBr + 6HN{O_3} \to 6KN{O_3} + 3B{r_2} + 3{H_2}O\]$

1 mole of $\[KBr{O_3}\]$ produces 3 moles of $\[B{r_2}\]$

Therefore,

1/3 moles of $\[KBr{O_3}\]$ produces 1 mole of $\[B{r_2}\]$

$\[0.0700 \times \frac{1}{3}\]$ moles of $\[KBr{O_3}\]$ produces 0.0700 moles of $\[B{r_2}\]$

Thus, the number of moles of  $\[KBr{O_3}\]$ required to produce 0.0700 moles of $\[B{r_2}\]$ can be given as:

$\[x = 0.0700 \times \frac{1}{3}\]$

$\[x = 0.0233\]$

Hence, the number of moles of $\[KBr{O_3}\]$ are required to prepare 0.0700 moles of $\[B{r_2}\]$ is $\[0.0233\]$.