Question

When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction caco3(s)→cao(s)+co2(g) what is the mass of calcium carbonate needed to produce 47.0 l of carbon dioxide at stp? express your answer with the appropriate units?

Answer 1

The mass of calcium carbonate that is needed to produce 47.0 l of Co2 at STP is 209.8 grams

calculation

CaCO₃ (s) → CaO(s) +Co₂

calculate the moles CO2 at STP

AT STP 1 mole of gas = 22.4 L

? moles = 47.0 L

by cross multiplication

=[ (1 moles x 47.0 l) / 22.4 L] =2.098 moles

use of mole ratio to calculate the moles of CaCO₃

CaCo₃:Co₂ is 1:1 therefore the moles of CaCo₃ is also = 2.098 moles

mass = moles x molar mass

The molar mass of CaCo₃ = 40 +12 +(16 x3) =100 g/mol

mass =2.098 moles x 100 g/mol =209.8 grams