Calcium carbonate reacts with HCl to give CaCl₂ and CO₂ according to the reaction:
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)
What mass of 20% impure CaCO₃ is required to react completely with 250 ml of 0.50 M HCl?
(1) 6.25 g
(2) 5 g
(3) 4.75 g
(4) 7.8 g
Answer (2)
To react completely with 250 ml of 0.50 M HCl, approximately 6.25 g of impure CaCO₃ is required, factoring in that the CaCO₃ is only 20% pure. This reflects calculations based on the moles of HCl and the mole-to-mole ratio in the balanced equation. The selected answer option is (1) 6.25 g.
;
To solve this problem, we need to follow these steps:
Identify the reaction details : The reaction is between calcium carbonate ( CaCO 3 ) and hydrochloric acid ( HCl ) to produce calcium chloride ( CaCl 2 ) , carbon dioxide ( CO 2 ) , and water ( H 2 O ) . The balanced chemical equation is:
CaCO 3 ( s ) + 2 HCl(aq) → CaCl 2 ( a q ) + CO 2 ( g ) + H 2 O(l)
Determine the moles of HCl : We have 250 mL of 0.50 M HCl.
Moles of HCl = Molarity × Volume (L)
= 0.50 M × 0.250 L = 0.125 moles of HCl
Use the stoichiometry of the reaction : The balanced equation shows that 2 moles of HCl react with 1 mole of CaCO₃. Therefore, the moles of CaCO₃ needed are:
0.125 moles HCl × 2 moles HCl 1 mole CaCO 3 = 0.0625 moles CaCO 3
Calculate the mass of pure CaCO₃ required : The molar mass of CaCO₃ is approximately 100.09 g/mol. Thus, the mass is:
0.0625 moles × 100.09 g/mol = 6.255625 g
Account for impurity : The CaCO₃ is 20% impure, which means the pure CaCO₃ makes up 80% of the mass. Let's denote the total mass of the impure sample as m . We have:
0.80 m = 6.255625 g
Solving for m :
m = 0.80 6.255625 g = 7.81953125 g
Choose the closest answer from the options : The mass of the 20% impure CaCO₃ required is approximately 7.8 g.
Thus, the correct answer is (4) 7.8 g .
