CHEM 1115 Summer 2018 Quiz 6 Problem S 9 of 16> When heated, calcium carbonate d
ID: 637916 • Letter: C
Question
CHEM 1115 Summer 2018 Quiz 6 Problem S 9 of 16> When heated, calcium carbonate decomposes to yield caldium oxide and carbon dioxide gas via the reaction CaCO,(s)?CaO(s) +CO,(s)what is the mass of cal um carbonate needed to produce 87.0 Lof carbon dioxide at STP? number of moles of a gas, n. The gas constant R 0.08206 L atm/(K mol) or 8.3145 J/(K mol). The equation can be rearranged as ollows to solve for na Py mass of CaCo,Vaue Units of the complete combustion of butane is2C,Hia(g) + 1302(g)+8C02(g)+10H,O()At 1.00 atm and 23 C, what is the volume of carbon dioxide formed by the combustion of 1.80 g of butane?Explanation / Answer
Ans 9
Part a
At STP
Pressure P = 1 atm = 101325 Pa
Temperature T = 273.15 K
Volume V = 87 L
Moles of CO2
n = PV/RT
= (101325 Pa x 87 L x 1m3/1000 L) / (8.3145 J/mol·K x 273.15 K)
n = 3.8815 mol
Moles of CaCO3 = moles of CO2 consumed
Mass of CaCO3 = moles x molecular weight
= 3.8815 mol x 100 g/mol
= 388.15 g
Part b
Pressure P = 1 atm = 101325 Pa
Temperature T = 23 + 273.15 = 296.15 K
Moles of butane = mass/molecular weight
= 1.80g / (58.12 g/mol)
= 0.03097 mol
From the stoichiometry of the reaction
Moles of CO2 produced
= 8 mol CO2 x 0.03097 mol C4H10 / 2 mol C4H10
= 0.12388 mol
Volume of CO2 produced
V = nRT / P
= 0.12388 mol x 8.3145 J/mol·K x 296.15 K / 101325 Pa
= 0.0030105 m3 x 1000L/m3
= 3.01 L