The weight of MnO 2 and the volume of HCl of specific gravity 1.2 g mL -1 and 4 % nature by weight, needed to produce 1.78 L of Cl 2 at STP. The reaction involved is: MnO 4+4 HCl arrow MnCl 2+2 H 2 O + Cl 2

Question

The weight of MnO 2 and the volume of HCl of specific gravity 1.2 g mL -1 and 4 % nature by weight, needed to produce 1.78 L of Cl 2 at STP. The reaction involved is: MnO 4+4 HCl arrow MnCl 2+2 H 2 O + Cl 2 (A) 0.48 L (B) 0.24 L (C) 0.12 L (D) 0.06 L

Tardigrade Admin · Accepted Answer

MnO 2+4 HCl arrow MnCl 2+2 H 2 O + Cl 2 Moles 0.08 4 × 0.8 (1.78/22.4)=0.08 =0.32 ⇒ W MnO 2=0.08 × 87 g =6.96 g ((10 × 4 × 1.2/36.5)) × V L =0.32 ⇒ V L =0.24 L

Q. The weight of $M n O_{2}$ and the volume of $H Cl$ of specific gravity $1.2 g m L^{- 1}$ and $4%$ nature by weight, needed to produce $1.78 L$ of $C l_{2}$ at $STP$ . The reaction involved is:
$M n O_{4} + 4 H Cl \to M n C l_{2} + 2 H_{2} O + C l_{2}$

$0.48 L$

$0.24 L$

$0.12 L$

$0.06 L$

$M n O_{2} + 4 H Cl \to M n C l_{2} + 2 H_{2} O + C l_{2}$
Moles $0.084 \times 0.8 \frac{1.78}{22.4} = 0.08$
$= 0.32$
$\Rightarrow W_{M n O_{2}} = 0.08 \times 87 g = 6.96 g$
$(\frac{10 \times 4 \times 1.2}{36.5}) \times V_{L} = 0.32 \Rightarrow V_{L} = 0.24 L$

Q. The weight of MnO2​ and the volume of HCl of specific gravity 1.2gmL−1 and 4% nature by weight, needed to produce 1.78L of Cl2​ at STP. The reaction involved is: MnO4​+4HCl→MnCl2​+2H2​O+Cl2​

0.48L

0.24L

0.12L

0.06L