if you have one mole of CH3CH3(g) + 2 moles of Cl2 (g) and form one mole of C2H4Cl2 (l) + 2 moles of HCl (g)
assuming a 90% yield calculate the amount of each reactant needed to form 1.5 kg of product
Here is a problem I worked in detail in stoichiometry. It isn't the same problem but the process is the same.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To calculate the amount of each reactant needed to form 1.5 kg of product, we first need to determine the molar mass of the product C2H4Cl2.
The molar mass of C2H4Cl2 can be calculated as follows:
- Carbon (C) has a molar mass of 12.01 g/mol.
- Hydrogen (H) has a molar mass of 1.01 g/mol.
- Chlorine (Cl) has a molar mass of 35.45 g/mol.
Multiply the molar masses by the number of atoms in the formula:
C2: 2 x 12.01 g/mol = 24.02 g/mol
H4: 4 x 1.01 g/mol = 4.04 g/mol
Cl2: 2 x 35.45 g/mol = 70.90 g/mol
Summing these results:
24.02 g/mol + 4.04 g/mol + 70.90 g/mol = 99.96 g/mol
Therefore, the molar mass of C2H4Cl2 is approximately 99.96 g/mol.
Since the yield is given as 90%, the actual amount of product obtained, assuming 100% conversion, would be (90/100) * 1.5 kg = 1.35 kg.
Now, let's calculate the moles of C2H4Cl2 obtained from 1.35 kg:
Moles of C2H4Cl2 = (mass of C2H4Cl2 / molar mass of C2H4Cl2)
Substituting the values:
Moles of C2H4Cl2 = (1350 g / 99.96 g/mol) = 13.51 mol
Since the balanced chemical equation shows that 1 mole of CH3CH3 reacts to form 1 mole of C2H4Cl2, we would need 13.51 moles of CH3CH3 to produce 13.51 moles of C2H4Cl2.
Similarly, the balanced chemical equation shows that 2 moles of Cl2 react to form 1 mole of C2H4Cl2. Therefore, we would need (2 * 13.51) = 27.02 moles of Cl2 to produce 13.51 moles of C2H4Cl2.
Considering a 90% yield, we can determine the amount of each reactant needed by dividing the mole quantities by 0.9 (0.9 represents 90% yield):
Moles of CH3CH3 needed = 13.51 mol / 0.9 = 15.01 mol
Moles of Cl2 needed = 27.02 mol / 0.9 = 30.02 mol
In conclusion, to produce 1.5 kg of product with a 90% yield, you would need approximately 15.01 moles of CH3CH3 and 30.02 moles of Cl2.