Part A When heated, calcium carbonate decomposes to yield calcium oxide and carb
ID: 1070075 • Letter: P
Question
Part A
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 71.0 L of carbon dioxide at STP?
Part B
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
A)
at STP, 1 mol = 22.4 L
So,
moles of CO2 = 71/22.4 = 3.17 mol
from given chemical equation,
moles of CaCO3 = moles of CO2
= 3.17 mol
mass of CaCO3 = number of moles * molar mass
= 3.17 mol*100 g/mol
= 317 g
Answer: 317 g
B)
moles of butane = mass/molar mass
= 3.80 / (4*12 + 1*10)
= 3.80 / (58)
=0.0655 mol
from given reaction,
moles of CO2 = 4*moles of butane
= 4* 0.0655
= 0.262 mol
to find volume of CO2 use;
P*V = n*R*T
1*V = 0.262*0.0821 * (23 + 273)
V = 6.37 L
Answer: 6.37 L