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

Limiting Reagent or Reactant

For the react...

Question

For the reaction : $M n O_{2} + 4 H C l 100 % - -- \to M n C l_{2} + C l_{2} + 2 H_{2} O$ , $17.4 g M n O_{2}$ is dissolved in $500 m L$ of $H C l$ solution containing $7.3 g$ $H C l$ per litre. (molar mass of $M n = 55 g / m o l$ )

H C l

is the limiting reagent

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

M n O_{2}

is the limiting reagent

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

0.05 moles of

M n C l_{2}

will form

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

1120 mL

C l_{2}

gas will be liberated at

0^{o} C

and 1 atm

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

Open in App

Solution

The correct option is D 1120 mL $C l_{2}$ gas will be liberated at $0^{o} C$ and 1 atm
$M n O_{2} + 4 H C l 100 % - -- \to M n C l_{2} + C l_{2} + 2 H_{2} O$

moles of $M n O_{2} = \frac{17.4}{87} = 0.2$
moles of $H C l = \frac{7.3}{36.5} = 0.2$

$\frac{moles of M n O_{2}}{stoichiometric coefficeint} = \frac{0.2}{1} = 0.2$

$\frac{moles of H C l}{stoichiometric coefficeint} = \frac{0.2}{4} = 0.05$

$H C l$ is a limiting reagent.

4 moles of $H C l$ form 1 mole of $C l_{2}$
$n_{C l_{2}} formed from 0.2 mol of HCl = 0.05$
as molar volume at STP = $22.4 L$
$V_{C l_{2}} = 0.05 \times 22.4 = 1120 m L$
Similarly, $moles of M n C l_{2} = \frac{1}{4} moles of HCl$
$n_{M n C l_{2}} = 0.05$

Suggest Corrections

Similar questions

Q. For the reaction :

M n O_{2} + 4 H C l 100 % - -- \to M n C l_{2} + C l_{2} + 2 H_{2} O

17.4 g M n O_{2}

is dissolved in

500 m L

H C l

solution containing

7.3 g

H C l

per litre. (molar mass of

M n = 55 g / m o l

)

10

g of

M n O_{2}

on reaction with conc.

H C l

liberated

0.1

equivalent of

C l_{2}

(molecular weight of

M n = 55

g/mol). Hence, percentage purity of

M n O_{2}

is :

Q. If

0.80

mol of

M n O_{2}

and

146

g of

H C l

react, then :

M n O_{2} + 4 H C l ⟶ M n C l_{2} + C l_{2} + 2 H_{2} O

M n O_{2} + 4 H C l \to 2 H_{2} O + M n {C l}_{2} + {C l}_{2}

The reaction of

M n O_{2}

and

H C l

produces

2.24 L

{C l}_{2}

gas. What mass of

M n O_{2}

was used up by the reaction? (At. mass of

M n = 55 g / m o l

)

Q. A

0.5

g sample containing

M n O_{2}

on treatment with

H C l

liberates

C l_{2}

. The

C l_{2}

is passed into a solution of

K I

and

30.0

mL of

0.1 M N a_{2} S_{2} O_{3}

are required to titrate the liberated iodine. Calculate the percentage of

M n O_{2}

in the sample (as nearest integer).

[Given that the atomic weight of

M n = 55

g/mol]

Join BYJU'S Learning Program

For the reaction : MnO2+4HCl100%−−−→MnCl2+Cl2+2H2O, 17.4 g MnO2 is dissolved in 500 mL of HCl solution containing 7.3 g HCl per litre. (molar mass of Mn=55 g/mol)

For the reaction : $M n O_{2} + 4 H C l 100 % - -- \to M n C l_{2} + C l_{2} + 2 H_{2} O$ , $17.4 g M n O_{2}$ is dissolved in $500 m L$ of $H C l$ solution containing $7.3 g$ $H C l$ per litre. (molar mass of $M n = 55 g / m o l$ )