The ideal gas law PV = nRT relates pressure P, volume V, temperature T, and numb

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. mole) or 83145 J/(K-mole) the equation can be rearranged as follows to solve for n n = PV/RT This equation is useful when dealing with gaseous reactions because stoichiometric calculations involve mole ratios When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction C_aCO_3 (s) rightarrow C_aO (s) CO_2 (g) What is the mass of calcium carbonate needed to produce 49.0 L of carbon dioxide at STP? Express your answer with the appropriate units. mass of C_aCO _3 - 210 g Butane, C_4H_10, is a component of natural gas that is used as fuel for cigarette lighters the balanced equation of the complete combustion of butane is 2C_4H_10 (g) + 13O_2 (g) rightarrow 8CO_2 (g)+ 10H_2O(t) AT 1.00 atm and 23 degree C, what is the volume of carbon dioxide formed by the combustion of 1.80got butane? Express your answer with the appropriate units, volume of CO_

Explanation / Answer

Part A:

Given,

V = 49L

At STP,

T = 273.15K

P = 1 atm

R= 0.08206L.atm/molK

Using Ideal gas law,

PV = nRT

n = PV/RT

n = 1atm*49L/0.08206L.atmmol-1K-1*273.15K

n = 2.186moles

molar mass of CaCO3 = 100.09g/m

Mass = Molar Mass *moles

Mass of CaCO3 = 100.09g/mol*2.186mols = 218.8g

Mass of CaCO3 =218.8g

Part B :

P = 1.00atm

T = 23C =23+273K = 296K

V= ?

R = 0.08206L.atm/molK

Mass of butane = 1.80g

Molar mass of butane = 58.12g/mol

Moles of butane = 1.80g/58.12g/mol

= 0.031moles

n = 0.031

from the balanced reaction,2 moles of bitane are present,

so n=2*0.031 = 0.062

PV= nRT

V=nRT/P

= 0.062mol*0.08206L.atmmol-1K-1*296K/1atm

V = 1.506L

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