8-96 L of Cl 2 gas which is at STP is mixed with 100 mL of 5 M Ca OH 2 which reacts as follows-2 Ca OH 2-2 Cl 2- Ca OCl 2- CaCl 2-2 H 2 OFind the number of chlorine atoms present in Ca OCl 2 that is formed- A- 0-5 N AB- 0-2 N AC- N AD- 0-4 N A

The correct option is D 0.4N_AWe know that, n_Cl_2=Volume(L) at STP22.4=8.9622.4=0.4 mol n_Ca(OH)_2=Molarity× volume (in L) n_Ca(OH)_2=5×1001000=0.5 mol Rati ...

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Byju's Answer

Standard XII

Chemistry

Equivalent Mass

8.96 L of Cl ...

Question

8.96 L of $C l_{2}$ gas which is at STP is mixed with 100 mL of 5 M $C a (O H)_{2}$ which reacts as follows:
$2 C a (O H)_{2} + 2 C l_{2} \to C a (O C l)_{2} + C a C l_{2} + 2 H_{2} O$

Find the number of chlorine atoms present in $C a (O C l)_{2}$ that is formed.

N_{A}

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0.5 N_{A}

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0.2 N_{A}

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0.4 N_{A}

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Solution

The correct option is D $0.4 N_{A}$
We know that,
$n_{C l_{2}} = \frac{Volume(L) at STP}{22.4} = \frac{8.96}{22.4} = 0.4 m o l$
$n_{C a (O H)_{2}} = M o l a r i t y \times v o l u m e (i n L)$
$n_{C a (O H)_{2}} = 5 \times \frac{100}{1000} = 0.5 m o l$
Ratio of the reactanting species is 1:1 in the balanced reaction. The one which has fewer moles will be the limiting reagent.
So $C l_{2}$ is the limiting reagent and we know that:
$\frac{n_{C l_{2} (c o n s u m e d)}}{2} = \frac{n_{C a (O C l)_{2} (p r o d u c e d)}}{1}$
$n_{C a (O C l)_{2}} = 0.2 m o l$
Number of Cl atoms in bleaching powder
$= 2 \times 0.2 \times N_{A} = 0.4 N_{A}$

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8.96 L of Cl2 gas which is at STP is mixed with 100 mL of 5 M Ca(OH)2 which reacts as follows: 2Ca(OH)2+2Cl2→Ca(OCl)2+CaCl2+2H2O Find the number of chlorine atoms present in Ca(OCl)2 that is formed.

8.96 L of $C l_{2}$ gas which is at STP is mixed with 100 mL of 5 M $C a (O H)_{2}$ which reacts as follows:
$2 C a (O H)_{2} + 2 C l_{2} \to C a (O C l)_{2} + C a C l_{2} + 2 H_{2} O$

Find the number of chlorine atoms present in $C a (O C l)_{2}$ that is formed.