Some Basic Concepts Of Chemistry Question 126
Question: Calcium carbonate reacts with aqueous HCl to give $ CaCl _2 $ and $ CO _2 $ according to the reaction,
$ CaCO _3(s)+2HCl(aq)\to $ $ CaCl _2(aq)+CO _2(g)+H _2O(l) $ In this reaction, 250 mL of 0.76 M HCI reacts with 1000 g of $ CaCO _3. $ Calculate the mass of $ CaCl _2 $ formed in the reaction.
Options:
A) 11.1 g
B) 10.54 g
C) 5.25 g
D) 2.45 L
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Molar mass of $ CaCO _3=40+12+3\times 16=100gmo{l^{-1}} $ Moles of $ CaCO _3 $ in 1000 g. $ {n _{CaCO _3}}=\frac{\text{Mass(g)}}{Molar,mass} $
$ {n _{CaCO _3}}=\frac{1000}{100g,mo{l^{-1}}}=10,mol $
$ \text{molarity=}\frac{Moles,of,\text{solute(HCl) }\times\text{ 100}}{Volume,of,solution} $
$ 0.76=\frac{n _{HCl}\times 1000}{250} $
$ n _{HCl}=\frac{0.76\times 250}{1000}0.19,mol $
$ \underset{1mol}{\mathop{CaCO _3(s)}}+\underset{2,mol}{\mathop{2HCl(aq)}},\to CaCl _2(aq)+ $ $ CO _2(g)+H _2O(l) $ According to the equation, 1 mole of $ CaCO _3 $ reacts with 2 moles of HCl
$ \therefore $ 10 moles of $ CaCO _3 $ will react with $ \frac{10\times 2}{1}=20 $ moles of HCI. But we have only 0.19 moles HCl, so HCl is the limiting reagent and it limits the yield of $ CaCl _2 $ . 0.19 moles of HCl will produce $ \frac{1\times 0.19}{2}=0.095,mol,CaCl _2. $ Molar mass of $ CaCl _2=40+(2\times ,35.5)=111,g,mo{l^{-}} $
$ \therefore $ 0.095 mole of
$ CaCl _2=0.095\times 111=10.54,g $
