1. stoichiometry questions a) A steel cylinder contains 5.00 mole of graphite (n

Question

1. stoichiometry questions

a) A steel cylinder contains 5.00 mole of graphite (neat carbon) and 5.00 moles of molecular
oxygen. The mixture is ignited and all the graphite reacts. Combustion produces a mixture
of carbon monoxide and carbon dioxide gases. After the cylinder has cooled to its original
temperature, it is found that the pressure of the cylinder has increased by 17.0%. Calculate
the mole fractions of all the gases in the nal mixture.

b) A chemist weighed out 5.14 g of a mixture containing unknown amounts of solid barium oxide
and solid calcium oxide and placed the sample in a 1.50 L
ask containing gaseous carbon
dioxide at 30.0C and 750 Torr. After the reaction was complete, solid barium carbonate and
solid calcium carbonate were formed. The pressure of carbon dioxide remaining was 230 Torr.
Calculate the mass percentages of calcium oxide and barium oxide in the original mixture.

Explanation / Answer

a)  we have , 5.00 mole of graphite (neat carbon) and 5.00 moles of molecular oxygen and the pressure of the cylinder has increased by 17.0%.

therefore total moles of gas = 5.00 mol of gas + 17% of 5.00 molof gas = 5.85 mol of gas

the reaction taking place are

C(s) + O2 --> CO2 and   2C(s) + O2 --> 2CO
therefore

At the end you have 1.70 mole CO and 4.15 mol of CO2+O2

mole fraction of CO = 0.29 (1.70 mol/5.85 mol)

mole fraction of O2+CO2 = 0.71 (4.15 mol/5.85 mol)

You will have O2 remaining because C is limiting

You know that you used 1.70 mole of C to produce 1.70 mol of CO (using 0.85 mol of O2) so you have only 3.30 mol

of C left to produce 3.30 mol CO2 (using 3.30 mol O2)

Moles of O2 left = 5.00 - 0.85 - 3.30 = 0.85

mole fraction O2 is = 0.85/5.85 = 0.145

mole fraction CO2 is = 3.30/5.85 = 0.565

mole fraction CO is = 0.290

b)

we have

T = 30C + 273 = 303K , P = 750 Torr x 0.00131579 = 0.9868 atm , V = 1.5 L

PV=nRT

n = PV/RT

n = [(0.9868 atm)(1.5L)] / [(0.0821L atm /K/mol)(303K)]

n = 0.0595 mol CO2

after the reaction we have P = 230 Torr X 0.00131579 = 0.3016 atm

n = PV/RT

n = [(0.3026 atm)(1.5L)] / [(0.0821 L atm/K/mol)(303K)]

n= 0.0183 mol CO2

therefore moles of CO2 consumed = 0.0595 - 0.0183 = 0.0412 mol
n = g CaO

5.14 - n = g BaO
and

n x (1 mol CaO/56.08 g CaO) = n / 56.08 mol CaO

(5.14 - n ) x (1 mol BaO / 153.33 g BaO) = (0.0335 - n / 153.33) mol BaO

0.0412 mol CO2 = n / 56.08 mol CaO + (0.0335 - n / 153.33) mol BaO

0.0412 = n / 56.08 + 0.0335 - n / 153.33

on solving

n = 0.68 g CaO

5.14 - 0.68 = 4.46 g BaO

therefore

we get

(0.68g CaO / 5.14g mixture) x 100 = 13.2% CaO

(4.46g BaO / 5.14g mixture) x 100 = 86.8% BaO

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