Question

reaction:2Br-(aq) + Cl2(aq)---2Cl-(aq)+ Br2(aq) is used for commercial preparation of bromine from its salts.if we have 50ml of 0.060M sol.of NaBr.what volume of 0.050M sol. of Cl2 is needed to react completely with Br- ions?

Answer 1

Rewriting the equation :

2 Br⁻ (aq) + Cl₂(g) - - - - - > 2 Cl⁻ (aq) + Br₂ (g)

Mole ratio is 2:1

Moles of Bromide ions :

(50 / 1000) × 0.060 = 0.003 moles

Moles of Chlorine is :

0.003 / 2 = 0.0015 moles

(0.0015 / 0.05) × 1000 = 30 cm³

Answer 2

Answer: The volume of $Cl_2$ required to react with $Br^-$ is 30 mL.

Explanation:

Molarity is defined as the number of moles present in one liter of solution.

Mathematically,

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$     .....(1)

We are given:

Molarity of NaBr or $Br^-$ = 0.06 mol/L

Volume of NaBr or $Br^-$ = 50 mL = 0.05 L    (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$0.06mol/L}=\frac{\text{Moles of }Br^-}{0.05L}}\\\\\text{Moles of }Br^-=0.003mol$

For the given chemical reaction:

$2Br^-(aq.)+Cl_2(aq.)\rightarrow 2Cl^-(aq.)+Br_2(aq.)$

By Stoichiometry of the reaction:

2 moles of bromine ions react with 1 mole of chlorine gas

So, 0.003 moles of bromine ions will react with = $\frac{1}{2}\times 0.003=0.0015mol$ of chlorine gas

Now, to calculate the volume of chlorine gas, we use equation 1.

Molarity of chlorine gas = 0.005 mol/L

Moles of chlorine gas = 0.0015 mol

Putting values in equation 1, we get:

$0.05mol/L}=\frac{0.0015mol}{\text{Volume of }Cl_2}\\\\\text{Volume of }Cl_2=0.03L=30mL$

Hence, the volume of $Cl_2$ required to react with $Br^-$ is 30 mL.