How many grams of 2-propanol are expected from the complete reaction of 111 g of propene?
To find the number of grams of 2-propanol (isopropyl alcohol) produced from the complete reaction of 111 g of propene, we need to follow these steps:
Step 1: Write and balance the chemical equation for the reaction.
The reaction between propene and water produces 2-propanol.
C3H6 + H2O -> C3H8O
Step 2: Calculate the molar mass of propene (C3H6) and 2-propanol (C3H8O).
- Molar mass of C3H6: (3 x atomic mass of carbon) + (6 x atomic mass of hydrogen)
= (3 x 12.01 g/mol) + (6 x 1.01 g/mol)
= 36.03 g/mol + 6.06 g/mol
= 42.09 g/mol
- Molar mass of C3H8O: (3 x atomic mass of carbon) + (8 x atomic mass of hydrogen) + (1 x atomic mass of oxygen)
= (3 x 12.01 g/mol) + (8 x 1.01 g/mol) + (1 x 16.00 g/mol)
= 36.03 g/mol + 8.08 g/mol + 16.00 g/mol
= 60.11 g/mol
Step 3: Convert the mass of propene to moles.
- Moles of propene = Mass of propene / Molar mass of propene
= 111 g / 42.09 g/mol
Step 4: Use the mole ratio from the balanced equation to find the moles of 2-propanol produced.
From the balanced equation: 1 mole of propene produces 1 mole of 2-propanol.
Therefore, the moles of 2-propanol = Moles of propene
Step 5: Convert moles of 2-propanol to grams.
- Mass of 2-propanol = Moles of 2-propanol x Molar mass of 2-propanol
= Moles of propene x Molar mass of 2-propanol
= Moles of propene x 60.11 g/mol
Now, substitute the calculated values:
- Mass of 2-propanol = (111 g / 42.09 g/mol) x 60.11 g/mol
Calculating:
- Mass of 2-propanol = 166.68 g
Therefore, the expected mass of 2-propanol produced from the complete reaction of 111 g of propene is 166.68 grams.