Answer:
a. the mole fraction of CO in the mixture of CO and O2.
mole fraction = moles of CO/ Total moles of the mixture
Mole fraction of CO = 10/(10+12.5)=0.444
b. Reaction - CO(g)+½O2(g)→CO2(g)
Stoichiometry: 1 mole of CO react with 0.5mole of O2 to give 1 mole of CO2
So given,
At a certain point in the heating, 3.0 mol CO2 is present. Determine the mole fraction of CO in the new mixture.
3mol of CO2 is produced from 3 mols of CO and 1.5mol of O2
This means that unused mols are : 7mols of CO and 11mols of O2
Total product mixture = 3 + 7 + 11 = 21mols
mole fraction of CO = 7/21 = 0.33
Answer:
1. Volume as STP = 755 L
2. Outside temperature = 255 K
3. Percentage yield = 70.5%
Explanation:
1. At STP, pressure = 101.3 kpa, temperature = 0°C or 273.15 K
Using the general gas equation :
P1V1/T1 = P2V2/T2
P1 = 620 kpa
V1 = 140 L
T1 = 37°C or (273.15 + 37) K = 310.15 K
P2 = 101.3 kpa
V2 = ?
T2 = 273.15 K
V2 = P1V1T2/P2T1
V2 = 620 × 140 × 273.15 / 101.3 × 310.15
V2 = 755 L
2. Using Charles' gas law:
V1/T1 = V2/T2
V1 = 2.5 L
T1 = 290 K
V2 = 2.2 L
T2 = ?
T2 = V2T1/VI
T2 = 2.2 × 290 / 2.5
T2 = 255 K
3. Equation of reaction : 2 Al + 3 CuSO4 ---> Al2 (SO4)3 + 3 Cu
From equation of the reaction, 2 moles of Al produces 3 moles of Cu
Molar mass of Al = 27 g; Molar mass of Cu = 63.5 g
2 moles of Al = 2 × 27 g = 54 g; 3 moles of Cu = 3× 63.5 = 190.5 g
54 g of Al produces 190.5 g of Cu
1.87 g of Al will produce 190.5/54 × 1.87 g of Cu = 6.60 g of Cu
Percentage yield = actual yield /theoretical yield × 100%
Percentage yield = 4.65/6.60 × 100%
Percentage yield = 70.5%
Depends on where you live but generally speaking it is either June or July
