You visited us 1 times! Enjoying our articles? Unlock Full Access!

Stoichiometric Calculations

What was the ...

Question

What was the percentage utilisation of lime ( $C a O$ )?

29.5 %

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

14.75 %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

59.0 %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

60.0 %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $29.5 %$
$F_{2} + 2 N a O H ⟶ \frac{1}{2} O_{2} + 2 N a F + H_{2} O$
$2 N a F + C a O + H_{2} O ⟶ C a F_{2} + 2 N a O H$
According to process,
$F_{2}$ $C a O$
$1$ mole $1$ mole
$\frac{2000}{38}$ $\frac{10, 000}{56}$
$= 52.63$ kmol $= 178.56$ kmol
So, $%$ of $C a O$ used $= \frac{52.63 \times 100}{178.56} = 29.5 %$ .

Suggest Corrections

Similar questions

F_{2} + 2 N a O H \to \frac{1}{2} O_{2} + 2 N a F + H_{2} O

2 N a F + C a O + H_{2} O \to C a F_{2} + 2 N a O H

C a F_{2}

F_{2} + 2 N a O H \to \frac{1}{2} O_{2} + 2 N a F + H_{2} O

2 N a F + C a O + H_{2} O \to C a F_{2} + 2 N a O H

C a F_{2}

C a O

C a O

200 k g

C a {C O}_{3}

C a {C O}_{3} \to C a O + {C O}_{2}

56 k g

C a O

Join BYJU'S Learning Program