Question

A 0.15 g sample of zinc ore was brought in solution as ZnCl2 and the zinc salt separated and treated with ferrocyanide solution as,
2K4[Fe(CN)6] + 3ZnCl2 K2Zn3[Fe(CN)6]2 + 6KCl
25 ml of 0.01 M ferrocyanide solution was required. Determine percentage of zinc in the ore. (Zn = 65)

Answer 1

Answer:

The percentage of zinc present in the ore is $16.2$ %·

Explanation:

The balanced chemical equation is given as,

$2\ K_{4}[Fe(CN)_{6}]\ +\ 3\ ZnCl_{2}$  →  $K_{2} Zn_{3}[Fe(CN)_{6}]_{2} \ +\ 6KCl$

Given that,

Mass of Zn ore in solution  $=$ $0.15g$

The Molar mass of Zn      $=$  $65\ gmol^{-1}$

To find out the $%$% of Zn ore,

First, find out the number of moles of Zn ore brought in the solution as$ZnCl_{2}$,

⇒ no. of moles of Zn ore as $ZnCl_{2}$     $=$  $\frac{mass\ of \ Zn \ ore\ taken}{Molar\ mass\ of\ ZnCl_{2} }$

⇒                                                           $= \ \ \frac{0.15}{136}$

⇒                                                           $=\ \ 1.1029$ × $10^{-3}$    

Next, find out the number of moles of ferrocyanide solution required

Given that,

Molarity of the solution $=\ \ 0.01M$

Volume         $=\ \ 25ml\ \ =\ \ 25$ × $10^{-3}L$

Therefore, by using the molarity equation,

No. of moles of ferrocyanide solution   $=$  M × $V_{L}$

⇒                                                              $=\ \ 0.01\ *\ 25\ *\ 10^{-3}$

⇒                                                              $=\ \ 2.5\ *\ 10^{-4}\ mol$

From the equation, it is clear that,

$2$ mol of  $K_{4}[Fe(CN)_{6}]$ needs $3$ mol of $ZnCl_{2}$·  SO,

$2.5\ *\ 10^{-4}$ mol of $K_{4}[Fe(CN)_{6}]$ needs  $=$   $\frac{2.5\ *\ 10^{-4\ }*\ 3 }{2}$

                                                                 $=\ \ 3.75\ *\ 10^{-4} mol$ of $ZnCl_{2}$

We know that,

    $1 mol$ of $ZnCl_{2}$  $=$  $65g$ of Zn

∴   $3.75\ *\ 10^{-4}$ mol of $ZnCl_{2}$ $=\ \ \frac{3.75\ *\ 10^{-4}\ *\ 65 }{1}$

⇒                                             $=\ \ 0.0243g$ of Zn

To determine the percentage,

We know that,

     mass %     $=\ \ \frac{mass\ of\ the\ solute}{mass\ of\ the\ solution} \ *100$

⇒  % of Zn in ore  $=\ \ \frac{mass\ of\ the\ Zn}{Total\ mass\ of\ the\ ore} \ *\ 100$

⇒                           $=\ \ \frac{0.0243}{0.15} \ *100$

⇒                           $=\ \ 16.2$ % of Zn

Hence, the percentage of Zn present in the ore is $16.2$ % ·

Answer 2

Given:

• Mass of Zinc = 0.15 g
• The molar mass of Zn = 65
• Balanced Chemical equation:
• $2K_4[Fe(CN)_{6}]+3ZnCl_2$ →→→→ $K_{2}Zn_3[Fe(CN)_6]_2$
• Molarity of the ferrocyanide solution  = 0.01 M
• Volume of ferrocyanide solution (V) = 25 ml = $25*10^{-3}L$

To Find:

• Percentage of Zinc in the ore.

Solution:

• To find out the percentage of Zinc in the ore first we should find out the number of moles of Zn in the zinc chloride solution.
• No: of moles of Zn in $ZnCl_2$ = $\frac{Mass of Zn}{Molar mass of ZnCl_2}$ = $\frac{0.15}{136}$ = $1.102*10^{-3}$
• Finding out the number of moles of ferrocyanide solution,
• No: of moles of ferrocyanide solution = $M*V_L$ = 0.01*25 * $10^{-3}$ = 0.25*$10^{-3}$ = 2.5*$10^{-4}$
• From the given balanced equation it is clear that 2 moles of $K_4[Fe(CN)_6]$ requires 3 moles of $ZnCl_2$
• So we get an equation, No: of moles of $K_4[Fe(CN)_6]$  =  $\frac{2.5*10^{-4}*3 }{2} = 3.75*10^{-4}$ mol of Zinc chloride.
• By the previous step, we can find 1 mol of $ZnCl_2$ = 65g of Zn
• So, No: of moles of Zinc Chloride = $\frac{3.75*10^{-4}*65 }{1}$ = $243.75*10^{-4} = 0.0243 g$  of Zn.
• To Find the percenatge of Zn in the ore we have a formula,
• mass percentage = $\frac{mass of solute}{mass of solution} *100$ = $\frac{0.0243}{0.15}*100$ = 0.162*100 = 16.2%
• Mass percentage = 16.2%

The percentage of Zn in the ore = 16.2%