To calculate the weight of oxygen needed to completely combust 2000 grams of isopropyl alcohol (isopropanol, C3H8O), we first need to look at the balanced combustion equation:
2 C3H8O + 9 O2 → 6 CO2 + 8 H2O + 4012 kJ
From the equation, we see that 2 moles of isopropyl alcohol react with 9 moles of oxygen (O2).
Next, we need to determine how many moles of isopropyl alcohol are in 2000 grams. The molecular weight of isopropyl alcohol is approximately 60.1 g/mol:
Molar mass of C3H8O = (3 * 12.01) + (8 * 1.008) + (1 * 16.00) = 60.1 g/mol
Now, we can find the number of moles in 2000 grams:
Number of moles = mass (g) / molar mass (g/mol) = 2000 g / 60.1 g/mol ≈ 33.19 moles
According to the balanced equation, for every 2 moles of isopropyl alcohol, 9 moles of oxygen are consumed. To find out how many moles of oxygen are needed for 33.19 moles of isopropyl alcohol, we set up a proportion:
(9 moles O2 / 2 moles C3H8O) = (x moles O2 / 33.19 moles C3H8O)
Solving for x gives us:
x = (9 moles O2 / 2 moles C3H8O) * 33.19 moles C3H8O ≈ 149.86 moles O2
Now, we need to calculate the mass of oxygen needed. The molar mass of oxygen (O2) is about 32.00 g/mol:
Mass of O2 = number of moles * molar mass = 149.86 moles * 32.00 g/mol ≈ 4799.52 g
Thus, to completely combust 2000 grams of isopropyl alcohol, approximately 4799.52 grams of oxygen is required.