Sea water contains 65 xx10^-3 "g/L" of bromide ions. If all the bromide ions are converted to produce Br_2,how much sea water is needed to prepare 1 kg of Br_2? please tell the method.

May 26, 2017

$15384.6156003 L$ of sea water

Explanation:

We know that the sea water contains $\frac{65 \times {10}^{-} 3 \text{g of bromide ions}}{L}$

If you want $B {r}_{2}$ the following reaction takes place

Br^-) + Br^-) = Br_2 + 2e^-

Well this reaction cannot take place but this can be a half reaction
But this reaction can take place

$2 \text{HBr" +"H"_2"SO"_4 rightleftharpoons "Br"_2 + "SO"_2 + 2"H"_2"O}$

So this like stoichiometry

1mol of Br- reacts with 1mol of Br- to form 1 mol of $B {r}_{2}$
or 2mol of ${\text{Br}}^{-}$ react to form 1 mol of $B {r}_{2}$

First calculate the amount of moles of ${\text{Br}}_{2}$ formed when $65 \times {10}^{-} 3 \text{g of bromide ions}$ react

Recall

"grams of substance"/"molar mass of substance" = "moles"" "(1)

$\frac{65 \times {10}^{-} 3 g}{{\text{molar mass of Br}}^{-}}$

Since the Br- has an extra electron

$\text{Molar mass of Br"^-) = "Molar mass of Br" + "Molar mass of electron}$

$\frac{79.904 g}{m o l} + \frac{5.485 799 090 70 \cdot {10}^{-} 10 g}{\text{mol" = "79.9040000005g/mol}}$

Now plug the variables equation (1)

$\frac{65 \times {10}^{-} 3 g}{79.9040000005} = 0.00081347617 \text{mol}$

The mole ratio of bromine ions reacted to ${\text{Br}}_{2}$ formed is

$1 m o l : 0.5 m o l$

Therefore solving for ratio

$0.00081347617 : x = 1 : 0.5$

$0.00081347617 : \frac{1}{2} \cdot 0.00081347617 = 1 : 0.5$

0.00040673808 mol of ${\text{Br}}_{2}$ is formed

We need 1kg of ${\text{Br}}_{2}$ so we have to find the no. of moles in 1kg of ${\text{Br}}_{2}$

Rearranging equation 1 we get

$\text{grams of substance" = "moles" xx "molar mass}$

$1000 g = \text{molar mass of Br} \times 2 \times x$

$79.904 \cdot 2 x = 1000$

$159.808 x = 1000$

$\text{moles} = \frac{1000}{159.808}$

$= \text{6.25750901081moles}$

If $\left(\text{0.00040673808 mol of Br"_2)/("1L of sea water}\right)$

Then

$\left(\text{6.25750901081moles of Br"_2)/("x") = ("0.00040673808 mol of Br"_2)/("1L of sea water}\right)$

$x$ is the desired amount of sea water

$6.25750901081 = 0.00040673808 x$

$x = \frac{6.25750901081}{0.00040673808}$

$15384.6156003 L$