How many grams of oxygen are needed to burn 22.8 grams of C8H18?

80.0 g
240. g
40.0 g
20.0 g
120. g

depends on the reaction. Is it

2C8H18 + 9O2 = 16C + 18H2O ?

If so, since you have 22.8/114 = 0.2 moles of C8H18, you will need 9/2*.2 = 0.9 moles of O2, or 28.8g

nope. So, Fix the reaction and adjust the moles of O2 needed.

To determine the number of grams of oxygen needed to burn 22.8 grams of C8H18, we need to balance the chemical equation for the combustion reaction between C8H18 (octane) and oxygen (O2).

The balanced equation for the combustion of octane is:
C8H18 + 12.5 O2 -> 8 CO2 + 9 H2O

From the balanced equation, we can see that for every 1 mole of octane (C8H18), we need 12.5 moles of oxygen (O2).

First, we need to calculate the number of moles of octane (C8H18) in 22.8 grams using its molar mass. The molar mass of octane is calculated by adding up the atomic masses of the elements in its chemical formula (12.01 g/mol * 8 + 1.01 g/mol * 18).

Molar mass of octane (C8H18) = 114.22 g/mol

Moles of octane (C8H18) = Mass of octane (22.8 g) / Molar mass of octane (114.22 g/mol)

Next, we can use the mole ratio from the balanced equation to determine the moles of oxygen (O2) needed.

Moles of oxygen (O2) = Moles of octane (C8H18) * 12.5

Finally, we can convert the moles of oxygen (O2) to grams using the molar mass of oxygen (O2) (32.00 g/mol).

Mass of oxygen (O2) = Moles of oxygen (O2) * Molar mass of oxygen (32.00 g/mol)

By substituting the calculated values into the formula, we find:
Mass of oxygen (O2) = (22.8 g / 114.22 g/mol) * 12.5 * 32.00 g/mol

Simplifying the equation, we get:
Mass of oxygen (O2) = (22.8 * 12.5 * 32.00) / 114.22 g ≈ 80.0 g

Therefore, the answer is 80.0 grams of oxygen (O2) are needed to burn 22.8 grams of C8H18 (octane).