Part A When heated, calcium carbonate decomposes to yield calcium oxide and carb

Question

Part A When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction The ideal gas law CaCOs(s) Cao(s)CO2(g) relates pressure P, volume V, temperature T, and number of moles of a gas, ra. The gas constant R equals 0.08206 L atm/(K mol) or 8.3145 J/(CK mol). The equation can be rearranged as What is the mass of calcium carbonate needed to produce 27.0 L of carbon dioxide at STP? Express your answer with the appropriate units. View Available Hint(s) follows to solve for : RT This equation is useful when dealing with gaseous reactions because stoichiometric calculations involve mole ratios. mass of CaCOVa Units Submit Part B Butane, CaHi0, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the compiete combustion of butane is 2C4Hio(g)+1302 (9)+8C02 (9)+ 10H200) At 1.00 atm and 23 C, what is the volume of carbon dioxide formed by the combustion of 3.60 g of butane? Express your answer with the appropriate units View Available Hint(s)

Explanation / Answer

part -A

the equation

CaCO3 (s) = CaO(s) + CO2 (g)

1 mol of CaCO3 produce 1 mol of  CO2

at STP 1 mol of gas = 22.4 L

mol of co2 in 27.0 L = 27.0 / 22.4 = 1.205 mol co2

this will require 1.205 mol  CaCO3

molar mass of  CaCO3 = 100 g/mol

1.205 mol = 1.205 * 100 = 120.5 g  CaCO3

part - B

the equation

2 C4H10 (g) + 13 O2(g) = 8CO2(g) + 10 H2O (l)

we have 2 : 8 or ( 1 : 4) ratio between C4H10 and CO2 this means that for every mole of C4H10 used in the reaction, 4 moles of CO2 will produce

molar mass of C4H10 = 58 g/mol

n butane = m / molar mass = 3.60 / 58 = 0.062 moles

thus we got n CO2 = 4 * nbutane = 4 * 0.062 = 0.248 moles

ideal gas equation

PV = nRT

V = nRT / P

V = 0.248 *0.082 * 298.15 / 1 .00

V = 6.063 L

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