How many grams of oxygen gas are needed to react with 19.0 mL of acetone?

Question

 Acetone has a density of 0.786 g/mLg/mL.

Answer 1

First, it is important to write down the balance equation for the combustion of acetone:

C3H6O + 4 O2 -> 3 CO2 + 3 H2O

To find the grams of oxygen gas needed to react with 19.0 mL of acetone, we need to follow these steps:

1. Calculate the moles of acetone:
2. Calculate the moles of oxygen gas:
3. Calculate the mass of oxygen gas:

1. Calculating moles of acetone:
2. Step 1.1: Find the acetone's density: The acetone's density is 0.791 g/mL.
3. Step 1.2: Calculate the mass of acetone: 19.0 mL * 0.791 g/mL = 15.03 g
4. Step 1.3: Calculate the moles of acetone: Molar mass of acetone (C3H6O) = 58.08 g/mol moles of acetone = mass / molar mass = 15.03 g / 58.08 g/mol = 0.259 mol
5. Calculate the moles of oxygen gas:
6. The balanced chemical equation shows that 1 mole of acetone reacts with 4 moles of oxygen gas. Therefore, we need four times the number of moles of acetone to find the number of moles of oxygen gas:
7. moles of oxygen gas = 0.259 mol acetone * 4 mol O2/mol acetone = 1.036 mol O2
8. Calculate the mass of oxygen gas:
9. Molar mass of oxygen gas (O2) = 32.00 g/mol mass of oxygen gas = moles * molar mass = 1.036 mol * 32.00 g/mol = 33.152 g

Therefore, 33.152 grams of oxygen gas must react with 19.0 mL of acetone.