± Stoichiometric Relationships with Gases Part A When heated, calcium carbonate

Question

± Stoichiometric Relationships with Gases Part A When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction What is the mass of calcium carbonate needed to produce 57.0 L of carbon dioxide at STP? Express your answer with the appropriate units The ideal gas law CaCO3 (s)CaO (s) + CO2 (g) 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 Hints PV RT This equation is useful when dealing with gaseous reactions because stoichiometric calculations involve mole ratios. mass of CaCO,-Vale g mass of CaCO3 = Submit My Answers Give Up 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 2C 4H10 (9)-1302 (9)-8CO2 (g) + 10H20 (I) At 1.00 atm and 23 °C, what is the volume of carbon dioxide formed by the c Express your answer with the appropriate units of 1.60 g of butane? Hints volume of CO2-1 Value L Submit My Answers Give Up

Explanation / Answer

Part A

at STP 1 mole gas occupy volume = 22.414 L then 57 liter volume of carbon dioxide =

57 / 22.414 = 2.543 mole

57 liter of CO2 at STP = 2.543 mole

According to reaction 1 mole of CaCO3 produce 1 mole of CO2 then to produce 2.543 mole of CO2 require 2.543 mole of CaCO3

molar mass of CaCO3 = 100.0869 gm/mole then 2.543 mole of CaCO3 = 2.543 X 100.0869 = 254.526 gm

to produce 57 L CO2 required CaCO3 = 254.526 gm

part B

molar mass of Butane = 58.12 gm/mole then 1.60 gm of butane = 1.60 / 58.12 = 0.02753 mole

According to reaction 2 mole of butane produce 8 mole of CO2 then 0.02753 mole of butane produce CO2 =

8 X 0.02753 / 2 = 0.11 mole of CO2

now use ideal gas equation to calculate volume of CO2

We know that PV = nRT

V = nRT/P

n = 0.11 mole,

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

P= 1.00 atm,

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

V = ?

Substitute these value in above equation.

V = 0.11 X 0.08205 X296.15 / 1 = 2.68 L

2.68 L CO2 produced.

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