The ideal gas law PV=nRT relates pressure P, volume V, temperature T, and number
ID: 958291 • Letter: T
Question
The ideal gas law PV=nRT relates pressure P, volume V, temperature T, and number of moles of a gas, n. The gas constant R equals 0.08206 Latm/(Kmol) or 8.3145 J/(Kmol). The equation can be rearranged as follows to solve for n: n=PVRT This equation is useful when dealing with gaseous reactions because stoichiometric calculations involve mole ratios.
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 53.0 L of carbon dioxide at STP? Express your answer with the appropriate units.
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? Express your answer with the appropriate units.
Explanation / Answer
Part A
1 mol of any can can occupy 22.4 L at STP . it is the key point to solve this proble
CaCO3 -------------------------> CaO + CO2
1mol 22.4 L
x mol 53.0 L
1 mol CaCo3 produce ----------------> 22.4 L CO2 at STP
x mol CaCO3 produce ----------------> 53.0 L CO2 at STP
x = 53.0 / 22.4
x = 2.37 moles
mass of CaCO3 = moles x molar mass = 2.37 x 100 = 237 g
mass of CaCO3 = 237 g
Part B
butane moles = 3.80 / 58 = 0.066
2C4H10(g) + 13O2(g)------------->8CO2(g) + 10H2O(l)
2 mol 8mol
0.066 mol ???
mole of CO2 produced = 8 x 0.066 / 2 = 0.262
T = 23 + 273 = 296 K
P = 1atm
P V = n R T
1 x V = 0.262 x 0.0821 x 296
V = 6.37 L
volume of CO2 = 6.37 L
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