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ADNAN MUHAMMED

Grade 12,

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112 L Cl 2 gas at STP is passed in 10 L KOH solution, containing 1 mole of potassium hydroxide per liter. Calculate the total moles of KCI produced, rounding it off to nearest whole number. (Yield of chemical reactions are written above the arrow ( →) of respective reaction)

112 L Cl2 gas at STP is passed in 10 L KOH solution, containing 1 mole of potassium hydroxide per liter. Calculate the total moles of KCI produced, rounding it off to nearest whole number. (Yield of chemical reactions are written above the arrow ( →) of respective reaction)

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Grade:11

1 Answers

Arun
25750 Points
3 years ago
Given equations are;
 
 
Cl2 + 2 KOH --------> KCl + KClO + H2O
 
3 KClO --------> 2 KCl + KClO3
 
4 KClO3 --------> 3 KClO4 + KCl
 
 
Before looking into the 1st equation,
 
we know,
 
22.4 litres of Cl2 indicates 1 mole of it
 
so, 4.48 litres of Cl2 indicates 0.2 moles
 
 
Also, number of moles of KOH = given mass/molar mass
 
= 11.2/56
 
= 0.2 moles
 
 
The 1st given equation is
 
Cl2 + 2 KOH --------> KCl + KClO + H2O
 
 
According to this equation, each mole of Cl2 requires 2 moles of KOH.
 
So, 0.2 moles of Cl2 requires 0.4 moles of KOH.
 
 
But there is only 0.2 moles of KOH present in the sample,
 
So, Only 0.1 moles of Cl2 will be able to participate in the reaction consuming all the 0.2 moles of KOH.
 
 
The other 0.1 moles of Cl2 left, will not have any KOH to react with! Hence it is the excess reagent which is left behind

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