2 C8H18(g) + 25 O2(g)  16 CO2(g) + 18 H2O­(g) at STP

How many moles of O2 are needed to react with 60.0g of C8H18 (octane)?

mole of 02 = 60/114 = 53 mols

2 moles of octane =25 mols of 02

0.53/2 x 25= 6.625

answer is 6.6 mols

How many litres of CO2 are produced?

16 x 0.53mol/2mol = 4.24 moles

4.24mol x 22.4 L = 94.976

ANSWER = 95.0 L of co2

Using 52.6 and not 53, that is

60.0/114 x (16/2) x 22.4 = 94.3157 which rounds to 94.3 L to 3 S.F.

60/0/114 = 52,6 and not 53

so 60.0/114 x (25/2) = 6.5789 mols. You're allowed 3 s.f. (60.0 has 3) so round to 6.58 mols O2 needed.

How many litres of CO2 are produced?

16 x 0.53mol/2mol = 4.24 moles

4.24mol x 22.4 L = 94.976

ANSWER = 95.0 L of co2

C8H18(l) + 25 O2(g) ----> 16 CO2(g) + 18 H2O(g) Suppose we burn 24 moles of C8H18, how many moles of CO2 form?

To find out how many moles of O2 are needed to react with 60.0g of C8H18 (octane), you first need to calculate the number of moles of C8H18. You can do this by dividing the mass of C8H18 (60.0g) by its molar mass.

The molar mass of C8H18 can be determined by adding up the molar masses of carbon (C) and hydrogen (H). The molar mass of carbon is 12.01 g/mol, and the molar mass of hydrogen is 1.01 g/mol. Since there are 8 carbon atoms and 18 hydrogen atoms in octane, the molar mass of octane is:

(8 * 12.01 g/mol) + (18 * 1.01 g/mol) = 114.22 g/mol

Next, divide the mass of octane (60.0g) by its molar mass:

60.0 g / 114.22 g/mol = 0.5259 mol

Now you can use the balanced equation to determine the molar ratio of O2 to octane. From the balanced equation, you can see that 2 moles of octane reacts with 25 moles of O2.

So, for 0.5259 moles of octane, you can calculate the moles of O2 required using the ratio:

(0.5259 mol octane / 2 mol octane) * (25 mol O2 / 1 mol octane) = 6.574 moles of O2

Rounding to the appropriate number of significant figures, you get approximately 6.6 moles of O2.

To calculate the volume of CO2 produced, you first need to determine the number of moles of CO2 formed. Using the balanced equation, you can see that for every 2 moles of octane reacted, 16 moles of CO2 are produced.

So, for 0.5259 moles of octane, you can calculate the moles of CO2 produced using the ratio:

(0.5259 mol octane / 2 mol octane) * (16 mol CO2 / 1 mol octane) = 4.238 mol CO2

Finally, to convert moles of CO2 to liters, you can use the ideal gas law which states that at STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 liters.

Therefore, the volume of CO2 produced is:

4.238 mol CO2 * 22.4 L/mol = 94.976 L

Rounding to the appropriate number of significant figures, you get approximately 95.0 L of CO2.

C8H18(l) + 25 O2(g) ----> 16 CO2(g) + 18 H2O(g) This time we use 40 moles of O2 to combust, how many moles of H2O form?