Potassium chlorate (KCIO4) is made in the following sequence of reactions:
Cl2(g) + KOH(cold & diluted) —>KCI+KCIO+H2O
KCIO -> KCl + KCIO3
KCIO3 →KCIO4 +KCI
What mass of Cl2 is needed to produce 1.385 kg of KCIO4?
For the above set of reactions what will be amount of KCl produced when 142 g of Cl2 is taken?
Answers
Answer:
The arrangement of conditions includes;
Cl
2
+2KOH→KCl+KClO+H
2
O......(1)
3KClO→2KCl+KClO
3
......(2)
4KClO
3
→3KClO
4
+KCl.......(3)
From the eqn(3)
The proportion
KCl
KClO
4
=
1
3
Implying that we have 3 moles of KClO
3
We have 1mole of KCl
Since the molar mass of KClO
4
is 138.55,
We have =
138.55
156
Moles delivered =1.126moles
On the off chance that we accept 100% yield these originated from (1.125×
3
4
) moles of KClO
3
=1.501 moles of KClO
3
required to deliver 156g of KClO
4
Moles of KClO expected to deliver 1.501 moles of KClO
3
=1.501×3=4.504 moles of KClO from eqn(2)
From eqn(1)
Moles of Cl2=moles of KClO =4.504 moles.
Along these lines, the mass of Cl
2
=4.504×71
Mass of Cl
2
=319g to deliver 156g of KClO
4
if 100% yield is accepted in the three compound responses.
Hence, this is the answer.
Explanation: