Question

calculate the milli moles of Br2 produced when 10 ml of 0.1 molar (BrO3) minus reacts with excess of Br -

Answer 1

Answer:  3 milli moles of bromine gas is produced when 10 ml of 0.1 molar bromate ions reacts with excess of bromide ions.

Explanation:

Molarity of bromate ion = 0.1 M

$Molarity=\frac{Moles}{volume(L)}$

$moles=0.1 mol/L\times 0.01 L=0.001 moles$

$BrO_2^-+6Br^-\rightarrow Br^-+3O^{2-}+3Br_2$

According to reaction, the  1 mol of bromate ion gives 3 moles of bromine gas.

Then 0.001 moles of bromate ion will give:

$\frac{3}{1}\times 0.001 =0.003 moles$ of $Br_2$

Milli moles of $Br_2$=3 milli moles

3 milli moles of bromine gas is produced when 10 ml of 0.1 molar bromate ions reacts with excess of bromide ions.

Answer 2

Answer:3

Explanation: