Question

Cl2 + 2 KOH = KCl + KClO + H2O
3 KClO = 2KCl + KClO3
4 KClO3 = 3KClO4 + KCl How much Cl2 is needed to prepare 400g of KClO4??? in grams Ive got 38.056 gm

Answer 1

yar m cal ni krungi show me ur claculus i ll chk ok

Answer 2

@srinidhijha i think you a very trivial error in the calculations regarding the placement of the decimal point. My answer reads something like 380.56 gms... (as you need approx. 12 moles of diChlorine gas for the reqd. amt. of KClO4/

Answer 3

kaise kiya??

Answer 4

its a wrong attempt...im goin \mad

Answer 5

apoorv show me ur calculation plz

Answer 6

mene aise koi qn ni kiya :P

Answer 7

arey calculate backwards okay? 3 moles kclo4 need 4 moles of kclo3. now prev. reacn. to make 4 moles of kclo3, you need 12 moles of kclo.
and for 12 moles of kclo, you need 12 moles of cl2.

so basically, for 3 mole of kclo4, i need 12 moles of cl2. that is no. of reqd. moles of cl2 will be 4 times that of kclo4 made.

now, 400gms kclo4, i.e. approx 3.9 moles is to me made. 4 times that no. of moles i.e. approx 11.6 moles of chlorine gas is needed. so wt. of 11.6 moles of chlorine gas shall be= 11.6 x 71 =810gms.

SORRY my earlier answer was WRONG. kindly check this one.

Answer 8

apoorv dont u think wen we doing bakward there will be 12 moles of Cl and 6moles of Cl2

Answer 9

nahi. give it a thought. see the whole reaction.

Answer 10

yar Cl2 mene dekha to 6 hi hain n apne 12 Cl ko dekh k likha h

Answer 11

kisi ne mera answer dekha kya isse pahle waala

Answer 12

arrreee yaaar

Answer 13

arey heena, see int he first thingy, 12 moles of Cl2 gas. okay? so we get 24 moles of Cl atoms. now out of that 12 are being used in producing KCl and 12 for KClO in the first reaction. so. so what?? that's what we need right????

BAZINGA PUNK!!

Answer 14

main to khali Cl pe laga rha hun
in the 3rd rxn
3 Cl(in KCLO4)---> 4 Cl = moles oof Cl in KClO3
==> 2.9 moles ke liye --> 2.89 x 4/3 = 3.85 moles of Cl in KClO3
in 2nd rxn
1 Cl in KClO3---> 3 Cl in KClO
=> for 3.85 Cl ---->3 x 3.85 Cl in KClO
in 1st reaction
for 1 Cl in KClO---> 2 Cl in CL2
=> for 3 x 3.85 Cl in KClO ---> 3.85 x3 x 2= 23.1moles of Cl = 35.5 x23.1= 820.05 g of Cl