Given:C8H18(l) + O2(g) ⟶ CO2(g) + H2O(g)How many grams of octane are needed to produce 985 grams of carbon dioxide?Write your final answer in two decimal places.Use the following mass numbers:O - 16C - 12H - 1

To solve this problem, we need to follow these steps:

1. First, we need to balance the chemical equation. The balanced equation is: 2C8H18(l) + 25O2(g) ⟶ 16CO2(g) + 18H2O(g)

2. Next, we calculate the molar mass of CO2 and C8H18. The molar mass of CO2 is (12.01 g/mol * 1) + (16.00 g/mol * 2) = 44.01 g/mol. The molar mass of C8H18 is (12.01 g/mol * 8) + (1.01 g/mol * 18) = 114.22 g/mol.

3. We then use the stoichiometry of the balanced equation to find out how many moles of C8H18 are needed to produce 1 mole of CO2. From the balanced equation, we can see that 2 moles of C8H18 produce 16 moles of CO2. Therefore, 1 mole of CO2 requires 2/16 = 0.125 moles of C8H18.

4. Now, we can calculate the number of moles of CO2 in 985 g. This is done by dividing the mass of CO2 by its molar mass: 985 g / 44.01 g/mol = 22.38 moles.

5. We multiply the number of moles of CO2 by the number of moles of C8H18 needed to produce 1 mole of CO2: 22.38 moles * 0.125 = 2.798 moles of C8H18.

6. Finally, we convert the number of moles of C8H18 to grams by multiplying by its molar mass: 2.798 moles * 114.22 g/mol = 319.54 g.

So, 319.54 grams of octane are needed to produce 985 grams of carbon dioxide.