Question

The minimum volume of HCl of specific gravity 1.2 and 3.65%(w/w) needed to produce 1.12L of Cl2 at 1 atm 273K

Mn02+HCL => MnCl2 +H2O +Cl2​

Answer 1

Answer:

166.6 mL

Explanation:

Balanced Equation:

$MnCl_2+4HCl ----> MnCl_2+2H_2O+Cl_2$

A mole of HCl produces 1 mole of Chlorine.

Number of moles in one ml of HCl solution.

Density of the solution is 1.2 g/mL

We have 3.65 %w/w.  

So in 1 mL there is 0.04385 g of HCl

Number of moles in one mL is 1.2*10^-3 mol/ml

Number of moles in 1.12 L of chlorine is 0.05 mol

(At STP gases occupy 22.4 L)

To produce 0.05 mol of chlorine we need 0.2 mol of HCl

So, volume of HCl we need is

$\frac{0.2}{1.2*10^{-3}}=166.6 mL$