Question

Acetone reacts with oxygen to form carbon dioxide and water according to the following reaction:

C₃H₆O_{(l, acetone)} + 4 O_{2 (g)} → 3 CO_{2 (g)} + 3 H₂O_(g)

If we start with 27.3 g of acetone and 28.0 g of oxygen, how many grams of carbon dioxide do we make?

Answer in 3 sig. figures please?

Answer 1

Number of moles of acetone = 27.3 g / 58.08 g/mol = 0.470 mole

Number of moles of O2 = 28.0 g / 32.0 g/mol = 0.875 mole

From the balanced equation we can say that

1 mole of acetone requires 4 mole of O2 so

0.470 mole of acetone will require

= 0.470 mole of acetone *(4 mole of O2 / 1 mole of acetone)

= 1.88 mole of O2

But we have 0.875 mole of O2 which is in short so O2 is limiting reactant

From the balanced equation we can say that

4 mole of O2 produces 3 mole of CO2 so

0.875 mole of O2 will produce

= 0.875 mole of O2 *(3 mole of CO2 / 4 mole of O2)

= 0.656 mole of CO2

mass of 1 mole of CO2 = 44.01 g

so the mass of 0.656 mole of CO2 = 28.9 g

Therefore, the mass of CO2 produced would be 28.9 g