Question

cement company uses 9.30 × 105 kg of limestone daily. The limestone decomposes upon heating, forming lime and carbon dioxide:

CaCO3(s) CaO(s) + CO2(g)

What volume of CO2 at 683 torr and 25.0°C is released into the atmosphere daily by this company?

Answer 1

The volume of CO₂ at 683 torr and 25.0°C released into the atmosphere daily by this company is 2.53×10⁸ L

How to determine mole of CaCO₃ that decomposed

• Mass of CaCO₃ = 9.30×10⁵ kg = 9.30×10⁵ × 1000 = 9.3×10⁸ g
• Molar mass of CaCO₃ = 100 g/mol

• Mole of CaCO₃ =?

Mole = mass / molar mass

Mole of CaCO₃ = 9.3×10⁸ / 100

Mole of CaCO₃ = 9.3×10⁶ moles

How to determine the moles of CO₂ produced

Balanced equation

CaCO₃ —> CaO + CO₂

From the balanced equation above,

1 mole of CaCO₃ reacted to produce 1 mole of CO₂

Therefore,

9.3×10⁶ moles of CaCO₃ will also react to produce 9.3×10⁶ moles of CO₂

How to determine the volume of CO₂

• Pressure (P) = 683 torr = 683 / 760 = 0.899 atm
• Temperature (T) = 25.0 °C = 25 + 273 = 298 K
• Gas constant (R) = 0.0821 atm.L/Kmol
• Number of mole (n) = 9.3×10⁶ moles

• Volume (V) =?

V = nRT / P

V = (9.3×10⁶ × 0.0821 × 298) / 0.899

V = 2.53×10⁸ L

Learn more about stoichiometry:

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