Question

12. (4 pts) One result of the increasing CO2 levels in the atmosphere is the increased acid content in the oceans, causing problems for sea life that make shells out of CaCO3. The following reaction is a simplified version of what happens to CaCO3 in an acidic environment. 2 HCl (aq) + CaCO3 (s) = CaCl2 (aq) + CO2 (g) + H2O(l) a. Calculate the volume of CO2 produced at a temperature of 14.2 °C and 1.43 atm when 10.4g of CaCO3 is consumed in the reaction. b. Calculate the grams of water produced in this reaction if 124 mL of CO2 are generated at a temperature of 18.4 °C and a pressure of 783 mmHg.

Answer 1

a)
Molar mass of CaCO3,
MM = 1*MM(Ca) + 1*MM(C) + 3*MM(O)
= 1*40.08 + 1*12.01 + 3*16.0
= 100.09 g/mol

mass of CaCO3 = 10.4 g
mol of CaCO3 = (mass)/(molar mass)
= 10.4/1.001*10^2
= 0.1039 mol


According to balanced equation
mol of CO2 formed = moles of CaCO3
= 0.1039 mol

Given:
P = 1.43 atm
n = 0.1039 mol
T = 14.2 oC
= (14.2+273) K
= 287.2 K

use:
P * V = n*R*T
1.43 atm * V = 0.1039 mol* 0.08206 atm.L/mol.K * 287.2 K
V = 1.7124 L
Answer: 1.71 L

b)
Given:
P = 783.0 mm Hg
= (783.0/760) atm
= 1.0303 atm
V = 124.0 mL
= (124.0/1000) L
= 0.124 L
T = 18.4 oC
= (18.4+273) K
= 291.4 K

find number of moles using:
P * V = n*R*T
1.0303 atm * 0.124 L = n * 0.08206 atm.L/mol.K * 291.4 K
n = 5.343*10^-3 mol

From reaction,
Mol of H2O produced = mol of CO2 produced
= 5.343*10^-3 mol

Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol

use:
mass of H2O,
m = number of mol * molar mass
= 5.343*10^-3 mol * 18.02 g/mol
= 9.626*10^-2 g
Answer: 9.63*10^-2 g