How many grams of water, H(subscript 2)O, and propene, C(subscript 3)H(subscript 6), can be formed from the reaction of 6.0 g of 2-propanol, C(subscript 3)H(subscript 8)O?

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C3H8O ==> C3H6 + H2O

mols C3H8O = grams/molar mass
Convert mols C3H8O to mols H2O using the coefficients in the balanced equation. (The ratio is 1:1).
Now convert mols H2O to grams.
g = mols H2O x molar mass H2O

To answer this question, we need to balance the chemical equation for the reaction of 2-propanol (C3H8O) to form water (H2O) and propene (C3H6). The balanced equation is:

C3H8O -> 2 C3H6 + 4 H2O

From the balanced equation, we can see that for every 1 mole of 2-propanol, we get 2 moles of propene and 4 moles of water.

Now, let's calculate the amount of substance (in moles) of 2-propanol, using its molar mass.

The molar mass of 2-propanol (C3H8O) is:
(3 * atomic mass of carbon) + (8 * atomic mass of hydrogen) + (1 * atomic mass of oxygen)
= (3 * 12.01 g/mol) + (8 * 1.01 g/mol) + (1 * 16.00 g/mol)
= 60.10 g/mol

Next, we can use the molar mass of 2-propanol to convert the given mass (6.0 g) into moles.

Number of moles of 2-propanol = mass / molar mass
= 6.0 g / 60.10 g/mol
= 0.0998 mol (approximately)

According to the balanced equation, for every 1 mole of 2-propanol, we get 2 moles of propene. Therefore, 0.0998 mol of 2-propanol will produce 2 * 0.0998 = 0.1996 mol of propene.

Similarly, for every 1 mole of 2-propanol, we get 4 moles of water. Therefore, 0.0998 mol of 2-propanol will produce 4 * 0.0998 = 0.3992 mol of water.

Finally, we can calculate the mass of water and propene using their respective molar masses.

Mass of water = number of moles * molar mass of water
= 0.3992 mol * 18.02 g/mol
= 7.19 g (approximately)

Mass of propene = number of moles * molar mass of propene
= 0.1996 mol * 42.08 g/mol
= 8.39 g (approximately)

Therefore, from the reaction of 6.0 g of 2-propanol (C3H8O), we can form approximately 7.19 g of water (H2O) and 8.39 g of propene (C3H6).