how many grams of oxygen are required to burn completely 570g of octane?

What is this so bad answer

To determine the number of grams of oxygen required to burn completely 570g of octane, we need to use the balanced chemical equation for the combustion reaction of octane.

The balanced equation for the combustion of octane is as follows:
2 C8H18 + 25 O2 → 16 CO2 + 18 H2O

From the equation, we can see that 2 moles of octane (C8H18) react with 25 moles of oxygen (O2) to produce 16 moles of carbon dioxide (CO2) and 18 moles of water (H2O).

To convert grams of octane to moles, we need to know the molar mass of octane:
(C8H18) = (12.01 g/mol * 8) + (1.01 g/mol * 18) = 114.22 g/mol

Now we can calculate the number of moles of octane:
570 g / 114.22 g/mol = 4.99 mol

Since the stoichiometric ratio between octane and oxygen is 2:25, we can use this ratio to find the moles of oxygen required:
4.99 mol of octane * (25 mol of O2 / 2 mol of octane) = 62.4 mol of O2

Finally, we can convert the moles of oxygen to grams by using the molar mass of oxygen (O2), which is 32 g/mol:
62.4 mol * 32 g/mol = 1996.8 g

Therefore, approximately 1996.8 grams of oxygen would be required to burn completely 570 grams of octane.

52 g

Write and balance the equation.

2C8H18 + 25O2 ==> 16CO2 + 18H2O

convert 570 gh C8H18 to mols. mol = g/molar mass octane.

Using the coefficients in the balanced equation convert mols octane to mols O2.

Now convert mols O2 to grams. g = mols x molar mass.