Question

8. Carbon dioxide can be made in the lab by the reaction of hydrochloric acid with calcium carbonate, shown in the equation below. How many liters of CO2 (g) at 30°C and 735 torr can be prepared from a mixture of 68.3 grams of CaCO3 (s) (100.09 g/mol) and 678 mL of 0.350 M HCI (10 pts)? R = 0.0821 Latm/molK CaCO3 (s) + 2 HCI (aq) ---> CaCl2 (aq) + H20 (1) + CO2 (9)

Answer 1

Given reaction is

CaCO₃(s) + 2HCl (aq)   → CaCl2 (aq) + H2O (l) + CO2

Calculation of moles

Number of moles of CaCO₃= Viven mass/ Molar Mass = 68,3g/(100.09gmol^-1 ) = 0.6824 mol

Number of moles of HCl = molarity * volume(L) = 0.350molL^-1  * 0.678L = 0.2373mol

According to the above equation, 1 mol CaCO₃ requires 2 mol of 2HCl, for the reaction.

Hence, for 0.6824 mol of CaCO₃(s) , the moles of HCl required would be 2 * 0.6824 mol = 1.3648 mol

But we have only 0.2373mol of HCl.

Thus, HCl is the limiting reagent .

So, CO₂(g) would be formed only from that amount of available HCl 0.2373mol

Since

2mol HCl gives CO₂ = 1mol

1mol HCl gives CO₂ = 1mol/2mol

0.2373mol HCl gives CO₂= 1mol * 0.2373mol /2mol = 0.11865mol

Hence mol of CO₂ formed is = 0.11865 mol

we using relation assuming form gas CO₂ is ideal

PV = nRT

V = nRT/P ..............................................................(1)

where Volume of gas = ?

n = number of mole of CO₂ = 0.11865mole

T = 30^o C = 303K

P = 735 torr = 735 torr * 0.00131579atm * torr^-1 = 0.96710565 atm

R = 0.0821 L*atm*Mol^-1K^-1

putting this value in equation number 1

V = 0.11865mole *  0.0821 L*atm*Mol^-1K^-1 * 303K /(0.96710565 atm) = 3.0519L

So the litres of CO₂ formed is 3.0519 L