Dilute hydrochloric acid reacts with calcium carbonate (CaCO3) to form calcium chloride (CaCl2), carbon dioxide (CO2), and water. The chemical equation involved is represented asCaCO3 + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)(Atomic Mass: Ca = 40, C = 12, O = 16, H = 1, Cl = 35.5)If 40.0 g of CaCO3 is treated with 40 g of HCl, then the mass of CO2 produced is?
The reaction is
CaCO 3 + 2HCl(aq)→ CaCl 2 (aq)+ H 2 O (l)+ CO 2 (g)
Molar mass of CaCO 3 is 100 g
Molar mass of HCl is 36.5 g mol-1
Molar mass of CO 2 is 44 g mol-1
100 g of CaCO 3 produces 44 g of CO2
40 g of CaCO 3 will produce (44 / 100)× 40
= 17.6 g of CO 2
2× 36.5 = 73 g of HCl ( 2 moles of HCl) produces 44 g of CO 2
40 g of HCl will produce (44 / 73)× 40
= 24.11 g of CO 2
Here, 40 g of HCl is producing 24.11 g of CO 2 and 40 g of CaCO 3 is producing 17.6 g of CO 2 . So, CaCO 3 is acting as limiting reagent. Amount of CO 2 produced will be 17.6 g.
CaCO 3 + 2HCl(aq)→ CaCl 2 (aq)+ H 2 O (l)+ CO 2 (g)
Molar mass of CaCO 3 is 100 g
Molar mass of HCl is 36.5 g mol-1
Molar mass of CO 2 is 44 g mol-1
100 g of CaCO 3 produces 44 g of CO2
40 g of CaCO 3 will produce (44 / 100)
= 17.6 g of CO 2
2
40 g of HCl will produce (44 / 73)
= 24.11 g of CO 2
Here, 40 g of HCl is producing 24.11 g of CO 2 and 40 g of CaCO 3 is producing 17.6 g of CO 2 . So, CaCO 3 is acting as limiting reagent. Amount of CO 2 produced will be 17.6 g.