Question

# The ideal gas law PV=nRT relates pressure P, volume V, temperature T, and number of moles...

The ideal gas law

PV=nRT

relates pressure P, volume V, temperature T, and number of moles of a gas, n. The gas constant Requals 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.

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 75.0 L of carbon dioxide at STP?

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 1.20 g of butane?

A) first calculate mole of carbon dioxide by using ideal gas equation

Ideal gas equation

PV = nRT             where, P = atm prrssure = 1 atm,

V = volume in Liter = 70 L

n = number of mole = ?

R = 0.08205L atm mol-1 K-1 =Proportionality constant = gas constant,

T = Temperature in K = 250C = 273.15+ 25 = 298.15 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (170) / (0.08205 298.15) = 2.86 mole

according to reaction to produce 1 mole of carbon dioxide require 1 mole calcium carbonate then to produce 2.86 mole of carbon dioxide require 2.86 mole of calcium carbonate

molar mass of calcium carbonate = 100.0869 gm/mol then 2.86 mole = 100.0869 2.86 = 286.25 gm

286.25 gm calcium carbonate require to produce 70 L carbon dioxide at STP

B) molar mass of butane = 58.12 gm/mol then 2 mole of butane = 116.24 gm

molar mass of carbon dioxide = 44.01 gm/mol then 8 mole carbon dioxide = 352.08 gm

According to reaction 2 mole of butane produce 8 mole of carbon dioxide that mean 116.24 gm butane produce 352.08 gm of carbon dioxide then 0.20 gm butane produce

352.08 1.20 / 116.24 = 3.63 gm of carbon dioxide

molar mass of carbon dioxide = 44.01 gm/mol then 3.63 gm = 3.63/ 44.01 = 0.0825 mole of carbon dioxide

caculate volume of carbon dioxide by using ideal gas equation

We know that PV = nRT

V = nRT/P

n = 0.0825 mole,

T = 230C = 23+273.15 = 296.15K,

P= 1 atm,

R = 0.08205 L atm mol-1 K-1 ( R = gas constant)

V = ?

Substitute these value in above equation.

V = (0.0825 0.08205 296.15)/ 1 = 2.0046 L

At 1.00 atm and 23 ∘C, 2.0046 L volume of carbon dioxide formed by the combustion of 1.20 g of butane

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