1. When heated, calcium carbonate decomposes to yield calcium oxide and carbon d

Question

1. 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 57.0 L of carbon dioxide at STP?

2. Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete combustion of butane is

2C4H10(g)+13O2(g)8CO2(g)+10H2O(l)

At 1.00 atm and 23 C, what is the volume of carbon dioxide formed by the combustion of 3.80 g of butane?

Explanation / Answer

1)
AT STP, molar volume in 22.4 L/mol
So,
number of mol of CO2 = volume / molar volume
= 57.0 L / 22.4 L/ mol
= 2.54 mol

from reaction,
moles of CaCO3 reacted = mol of CO2 formed
= 2.54 mol

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

use:
mass of CaCO3,
m = number of mol * molar mass
= 2.54 mol * 100.09 g/mol
= 2.54*10^2 g
Answer: 2.54*10^2 g

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