Question

Calculate the volume and no.of molecules of Co2 liberated at stp
if 100 gm. of Caco, is treated with Dil.HCI. Which contains
7.3 gm of dissolved HCl gas ?
Caco3+ 2 HCI → CaCl2(aq) + H2O + Co2
(Atomic mass of Ca = 40 u, C = 12 units,
0 = 16 units, H= 1 unit, CI= 35.5 U

Answer 1

Answer:

Volume of CO₂ = 2.24 L

No. of CO₂ molecules = 6.023 * 10²²

Explanation:

given:

$CaCO_3+ 2HCl --> CaCl_2 + H_2O + CO_2$

100 gm ( 1 mole ) of CaCO₃ reacts with 7.3 gm ( 0.2 mole ) of HCl

To find :

volume and no. of CO₂ liberated at STP

Here HCl is limiting agent .

so, 0.2 mole of HCl react with 0.1 mole of CaCO₃ and produces 0.1 mole of CO₂

0.1 mole of CO₂ = 4.4 grams of CO₂ = 6.023 * 10²² CO₂ molecules = 2.24 L of CO₂