100g of CaCO3 is mixed with 100g of HCl. If 740ml of a solution of NaOH is required to completely neutralize the excess HCl, the normality of this NaOH solution is?
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What is the solution to this: “100 g of CaCo3 is treated with 1L of 1N HCl. What would be the weight of CO2 liberated after the completion of the reaction?”
First, Lets write the balanced equation
CaCO3 + 2HCl -> CO2 + CaCl2 + H2O
It's clear that 1 mole of CaCO3 reacts with 2 miles of HCl to give 1 mole of CO2.
Now, according to question: 100g CaCO3= 1 mol of CaCO3
1N HCl=1M HCl (n factor for HCl is 1)
Molarity=No of moles/Volume in L
1=No. Of moles/1L
No. Of moles of HCl = 1
Thus, HCl is limiting Reagent.
Hence, only 0.5 moles of CaCO3 would react with 1 mole of HCl to give 0.5 mole of CO2.
Now, 0.5Moles of CO2= weight/molecular mass of CO2.
0.5=weight/44
Weight of CO2 taken= 22g
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