suppose 16.0g of benzene C6H6 is burned in oxygen.What mass of O2 is required for the complete combustion

suppose 16.0g of benzene C6H6 is burned in oxygen.What is the total mass of products expected from 16.04g of benzene

first we write the chemical equation:

C6H6 + O2 -> CO2 + H2O
then we balance this:
2C6H6 + 15O2 -> 12CO2 + 6H2O
then we solve for the molecular weight of C6H6. Note that mass of C = 12 and H = 1. thus,
6*12 + 6*1 = 78 g/mol
then we get the number of moles of C6H6 used in the reaction:
n = 16/78 = 0.205 moles

then we use stoich ratio from thethe equation to get moles of O2:
0.205 * (15 mol O2 / 2 mol C6H6) = 1.5375 mol O2
note: molecular mass of O2 = 32 g/mol
finally,
1.5375 * 32 = 49.2 g O2

Well, let's do some math!

To calculate the amount of O2 required for the complete combustion of benzene (C6H6), we need to know the balanced chemical equation for the reaction.

C6H6 + O2 → CO2 + H2O

According to the equation, we need one mole of benzene reacting with 15 moles of O2 to produce six moles of CO2 and three moles of H2O.

Now, we need to calculate the molar mass of benzene (C6H6).

The molar mass of carbon (C) is 12.01 g/mol, and there are six carbon atoms in benzene, so the carbon part contributes 6 × 12.01 = 72.06 g/mol.

The molar mass of hydrogen (H) is 1.01 g/mol, and there are six hydrogen atoms in benzene, so the hydrogen part contributes 6 × 1.01 = 6.06 g/mol.

Adding these two values together, we get a total molar mass of 78.12 g/mol for benzene (C6H6).

Since we have 16.0 g of benzene, we divide the given mass by the molar mass to find the moles of benzene:

16.0 g ÷ 78.12 g/mol ≈ 0.205 moles of benzene

According to the balanced equation, we find that the mole ratio between benzene and oxygen is 1:15. So, if we have 0.205 moles of benzene, we would need:

0.205 moles × 15 moles of O2/1 mole of benzene ≈ 3.075 moles of O2

Finally, we can calculate the mass of O2 required by multiplying the number of moles by its molar mass:

3.075 moles × 32.00 g/mol ≈ 98.4 g of O2

Therefore, approximately 98.4 grams of O2 would be required for the complete combustion of 16.0 grams of benzene. That's quite an appetite for O2!

To determine the mass of oxygen required for the complete combustion of benzene (C6H6), we first need to balance the chemical equation for the combustion reaction:

C6H6 + O2 -> CO2 + H2O

The balanced equation shows that 1 molecule of benzene reacts with 15 molecules of oxygen to produce 6 molecules of carbon dioxide and 3 molecules of water.

Now, let's calculate the molar mass of benzene (C6H6):
C = 12.01 g/mol
H = 1.01 g/mol

Molar mass of benzene (C6H6):
6(C) + 6(H) = (6 * 12.01 g/mol) + (6 * 1.01 g/mol) = 78.11 g/mol

Using the molar ratio from the balanced equation, we can calculate the amount of oxygen required:

Molar ratio of benzene to oxygen = 15:1

Mass of benzene = 16.0 g
Molar mass of benzene = 78.11 g/mol

Moles of benzene = Mass of benzene / Molar mass of benzene
= 16.0 g / 78.11 g/mol

To find the moles of oxygen required for complete combustion, we multiply the moles of benzene by the molar ratio:

Moles of oxygen = Moles of benzene x (Molar ratio of oxygen / Molar ratio of benzene)
= (16.0 g / 78.11 g/mol) x (1 mol O2 / 15 mol C6H6)

Now, let's calculate the mass of oxygen:

Mass of oxygen = Moles of oxygen x Molar mass of oxygen

The molar mass of oxygen is approximately 32.00 g/mol.

So, the final calculation is:

Mass of oxygen = (16.0 g / 78.11 g/mol) x (1 mol O2 / 15 mol C6H6) x 32.00 g/mol

Calculating this expression will give us the mass of oxygen required for the complete combustion of 16.0 g of benzene.

To find out the mass of oxygen (O2) required for the complete combustion of benzene (C6H6), we need to balance the chemical equation and use the stoichiometry.

The balanced chemical equation for the combustion of benzene is:
C6H6 + O2 -> CO2 + H2O

From the balanced equation, we can see that for every one molecule of benzene, we need 15 molecules of oxygen to react completely.

We start by calculating the molar mass of benzene:
C6H6 = (6*12.01 g/mol) + (6*1.01 g/mol) = 78.11 g/mol

Now we can calculate the number of moles of benzene:
moles of C6H6 = mass / molar mass
moles of C6H6 = 16.0 g / 78.11 g/mol = 0.205 mol

Since the molar ratio between C6H6 and O2 is 1:15, we can use this ratio to calculate the moles of O2 required:
moles of O2 = moles of C6H6 * (15 mol O2 / 1 mol C6H6)
moles of O2 = 0.205 mol * 15 = 3.075 mol

Finally, we can calculate the mass of O2 required:
mass of O2 = moles of O2 * molar mass of O2
mass of O2 = 3.075 mol * 32.00 g/mol = 98.4 g

Therefore, approximately 98.4 grams of oxygen (O2) are required for the complete combustion of 16.0 grams of benzene (C6H6).