The ideal gas law : P V = n R T relates pressure P , volume V , temperature 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

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 75.0 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. 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 75.0 L of carbon dioxide at STP

ideal gas law:

P*V = n*R*T

given that

P = 1 atm

V = 75.0 L of CO2 at STP

n CO2 = X mol of CO2

R = 0.082 L atm K^1 mol^1

T = 25 °C, convert from °C to K (273.15 + 25) = 298,15 K

X = P*V/R*T

X = 1*75/(0.082*298.15)

X = 75/24,45

X = 3,06 mol of CO2 at STP

MWCaCO3 = 100.09 = 1 mol
MWCO2 = 44.01 = 1 mol

MW = Molecula Weight (by the periodic table)

MWCaCO3 : MWCO2 = Y : (3.06*44.01)

100.09 : 44.01 = Y : 134,67

Y = (100.09*134.67)/44.01

Y =13,479.12 /44.01

Y = 306.27 = 306.3 g of calcium carbonate to produce 75.0 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?

2 C4H10 + 13 O2 --> 8 CO2 + 10 H2O
3.80 g / 58.124 g/mole = 0.065377 mole of butane
(8/2) x0.0 65377= 0.2615 mole of CO2
V = nRT/P
V = 0.2615 mole x 0.08206 x 296 K / 1.00 atm = 6.351 L

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.