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 L⋅atm/(K⋅mol) or 8.3145 J/(K⋅mol). 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.

Answer 1

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