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Standard XII

Chemistry

Concentration Terms - An Introduction (w/w etc)

Calculate the...

Question

Calculate the mass of pure $M n O_{2}$ to produce 35.5g of ${C l}_{2}$ according to the following reaction.
$M n O_{2} + 4 H C I ⟶ M n C I_{2} + {C l}_{2} + 2 H_{2} O$

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Solution

$35.5 g$ $C l_{2} \Rightarrow \frac{1}{2} m o l$ $C l_{2}$

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

To produce 0.5 mol $C l_{2}$ ,0.5mol $M n O_{2}$ required

$∴$ mass $= 0.5 \times 87 = 43.5 g$

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Similar questions

10 g

M n O_{2}

on reaction with HCI forms

2.24 L

{C I_{2}}_{(g)}

at NTP, the percentage impurity of

M n O_{2}

is:

M n O_{2} + 4 H C I \to M n C I_{2} + C I_{2} + 2 H_{2} O

calculate the mass of $M n O_{2}$ , needed to produce 1.78 litres of chlorine gas at STP according to the equation
$M n O_{2}$ + 4HCl $\to$ $M n C l_{2}$ + 2 $H_{2} O$ + $C l_{2}$

[H = 1, Cl = 35.5, Mn = 54.9]

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. The weight of

M n O_{2}

and the volume of

H C l

of specific gravity

1.2

g/mL and

4 %

nature by weight needed to produce

1.78

L of

C l_{2}

S T P

. The reaction involved is as follows:

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

Q. Chlorine is prepared in the laboratory by treating manganese dioxide

(M n O_{2})

with aqueous hydrochloric acid according to the reaction,

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

How many grams of HCl will react with 5 g

M n O_{2}

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Calculate the mass of pure MnO2 to produce 35.5g of Cl2 according to the following reaction.MnO2+4HCI⟶MnCI2+Cl2+2H2O

35.5g Cl2⇒12mol Cl2 MnO2+4HCl→MnCl2+Cl2+2H2O To produce 0.5 mol Cl2,0.5mol MnO2 required ∴ mass =0.5×87=43.5g

Calculate the mass of pure $M n O_{2}$ to produce 35.5g of ${C l}_{2}$ according to the following reaction.
$M n O_{2} + 4 H C I ⟶ M n C I_{2} + {C l}_{2} + 2 H_{2} O$

$35.5 g$ $C l_{2} \Rightarrow \frac{1}{2} m o l$ $C l_{2}$

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

To produce 0.5 mol $C l_{2}$ ,0.5mol $M n O_{2}$ required

$∴$ mass $= 0.5 \times 87 = 43.5 g$