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Preparation of Haloalkanes from Alcohol

29.4g of K2Cr...

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29.4g of K2Cr2O7 react with HCl K2Cr2O7 + HCl -> 2KCl + 2CrCl3 +3Cl2 +&H2O 1. What is the volume of CO2 evolved 2. How much HCl is required to completely consume K2Cr2O7

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Q. The equivalent mass of HCl in the given reaction is
k2cr2o7+14hcl gives 2kcl+2crcl3+3cl2+h2o

n

H C l

in the equation

K_{2} C r_{2} O_{7} + 14 H C l \to 2 K C l + 2 C r C l_{3} + 3 C l_{2} + 7 H_{2} O

Q. The equivalent weight of

H C l

in the given reaction is :

K_{2} C r_{2} O_{7} + 14 H C l \to 2 K C l + 2 C r C l_{3} + 3 C l_{2} + H_{2} O

Q. The equivalent mass of

H C l

in the following reaction

K_{2} C r_{2} O_{7} + 14 H C l \to 2 K C l + 2 C r C l_{3} + 3 C l_{2} + H_{2} O

6.13

K_{2} C r_{2} O_{7}

96 %

purity) is allowed to react with

320

H C l

solution having density

1.15

g/mL and containing

30 %

H C l

C l_{2}

K_{2} C r_{2} O_{7} + 14 H C l ⟶ 2 K C l + 3 C r C l_{3} + 7 H_{2} O + 3 C l_{2}

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Preparation of Haloalkanes from Alcohol

Standard XII Chemistry

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