| Calcium carbonate reacts with aqueous HCl to give \[CaC{{l}_{2}}\] and \[C{{O}_{2}}\] according to the reaction, |
| \[CaC{{O}_{3}}(s)+2HCl(aq)\to \]\[CaC{{l}_{2}}(aq)+C{{O}_{2}}(g)+{{H}_{2}}O(l)\] |
| In this reaction, 250 mL of 0.76 M HCI reacts with 1000 g of \[CaC{{O}_{3}}.\] Calculate the mass of \[CaC{{l}_{2}}\] formed in the reaction. |
A) 11.1 g
B) 10.54 g
C) 5.25 g
D) 2.45 L
Correct Answer: B
Solution :
| [b] Molar mass of |
| \[CaC{{O}_{3}}=40+12+3\times 16=100\text{ }g\text{ }mo{{l}^{-1}}\] |
| Moles of \[CaC{{O}_{3}}\] in 1000 g. |
| \[{{n}_{CaC{{O}_{3}}}}=\frac{\text{Mass(g)}}{\text{Molar}\,\text{mass}}\] |
| \[{{n}_{CaC{{O}_{3}}}}=\frac{1000}{100g\,mo{{l}^{-1}}}=10\,mol\] |
| \[\text{molarity=}\frac{\text{Moles}\,\text{of}\,\text{solute(HCl) }\!\!\times\!\!\text{ 100}}{\text{Volume}\,\text{of}\,\text{solution}}\] |
| \[0.76=\frac{{{n}_{HCl}}\times 1000}{250}\] |
| \[{{n}_{HCl}}=\frac{0.76\times 250}{1000}0.19\,mol\] |
| \[\underset{1mol}{\mathop{CaC{{O}_{3}}(s)}}\,\,+\underset{2\,mol}{\mathop{2HCl(aq)}}\,\to CaC{{l}_{2}}(aq)+\]\[C{{O}_{2}}(g)+{{H}_{2}}O(l)\] |
| According to the equation, |
| 1 mole of \[CaC{{O}_{3}}\] reacts with 2 moles of HCl |
| \[\therefore \]10 moles of \[CaC{{O}_{3}}\] 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 \[CaC{{l}_{2}}\]. |
| 0.19 moles of HCl will produce |
| \[\frac{1\times 0.19}{2}=0.095\,mol\,CaC{{l}_{2}}.\] |
| Molar mass of |
| \[CaC{{l}_{2}}=40+(2\times \,35.5)=111\,g\,mo{{l}^{-}}\] |
| \[\therefore \] 0.095 mole of |
| \[CaC{{l}_{2}}=0.095\times 111=10.54\,g\] |
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