Question

How many liters of carbon dioxide gas, measured at 20 degrees Celsius and 715 torr, are formed when 49 grams of chalk (CACO3, 100.9 g/mol) are reacted with 60 ml of 3.20 M sulfuric acid (H2SO4) according to the following molecular equation (R= .0821 Latm/molK).

CACO3 (s) + H2SO4 (aq) -> CO2 + H2O (l) + CASO4 (s)

Answer 1

Number of moles of CaCO3 , n = mass/molar mass

= 49 g / 100.9(g/mol)

= 0.486 moles

Number of moles of H2SO4 , n' = Molarity x volume in L

= 3.20 M x 60 mL x 10 ^-3 L/mL

= 0.192 moles

CaCO3 (s) + H2SO4 (aq) ----> CO2 + H2O (l) + CaSO4 (s)

1 mole of H2SO4 reacts with 1 mole of CaCO3

0.192 moles of H2SO4 reacts with 0.192 moles of CaCO3

So 0.486 - 0.192 moles of CaCO3 left unreacted so CaCO3 is excess reactant.

All of H2SO4 completly reacted it is limiting reactant.

Again 1 mole of H2SO4 produces 1 mole of CO2

0.192 mole of H2SO4 produces 0.192 mole of CO2

Calculation of volume of CO2 :

We know that ideal gas equation is PV = nRT

Where

T = Temperature = 20 oC = 20+273 = 293 K

P = pressure = 715 torr x ( 1 atm / 760 torr) = 0.941 atm

n = No . of moles = 0.192 moles

R = gas constant = 0.0821 L atm / mol - K

V= Volume of the gas = ?

Plug the values we get V = ( nRT) / P = 4.91 L