One of the compounds used to increase the octane rating of gasoline is toluene (pictured). suppose 43.3 ml of toluene (d = 0.867 g/ml) is consumed when a sample of gasoline burns in air. how many grams of oxygen are needed for complete combustion of the toluene? (a) how many grams of oxygen are needed for complete combustion of the toluene? g (b) how many total moles of gaseous products form? mol (c) how many molecules of water vapor form?

A. The mass of oxygen needed for complete combustion of the toluene is 117.52 g.

B. The total mole of gaseous products formed is 4.488 moles

C. The number of molecules of water vapor formed is 9.82×10²³ molecules

We'll begin by calculating the mass of toluene. This can be obtained as follow:

Volume = 43.3 mL

Density = 0.867 g/mL

Mass = Density × Volume

Mass of toluene = 0.867 × 43.3

C₆H₅CH₃ + 9O₂ —> 7CO₂ + 4H₂O

Molar mass of C₆H₅CH₃ = (6×12) + (1×5) + 12 + (3×1) = 92 g/mol

Mass of C₆H₅CH₃ from the balanced equation = 1 × 92 = 92 g

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mass of O₂ from the balanced equation = 9 × 32 = 288 g

SUMMARY:

From the balanced equation above,

92 g of C₆H₅CH₃ reacted with 288 g of O₂

A. Determination of the mass of O₂ needed for the complete combustion of the toluene

From the balanced equation above,

92 g of C₆H₅CH₃ reacted with 288 g of O₂.

Therefore,

37.5411 g of C₆H₅CH₃ will react with = (37.5411 × 288)/92 = 117.52 g of O₂

Thus, 117.52 g of O₂ is needed for the reaction.

B. Determination of the total number of mole of the gaseous products formed

Molar mass of C₆H₅CH₃ = 92 g/mol

Mass of C₆H₅CH₃ = 37.5411 g

Mole = mass / molar mass

Mole of C₆H₅CH₃ = 37.5411 / 92

C₆H₅CH₃ + 9O₂ —> 7CO₂ + 4H₂O

From the balanced equation above,

1 mole of C₆H₅CH₃ produced 7 moles of CO₂

Therefore,

0.408 mole of C₆H₅CH₃ will produce = 0.408 × 7 = 2.856 moles of CO₂

C₆H₅CH₃ + 9O₂ —> 7CO₂ + 4H₂O

From the balanced equation above,

1 mole of C₆H₅CH₃ produced 4 moles of H₂O

Therefore,

0.408 mole of C₆H₅CH₃ will produce = 0.408 × 4 = 1.632 moles of H₂O

Mole of CO₂ produced = 2.856 moles

Mole of H₂O produced = 1.632 moles

Total mole = 2.856 + 1.632

Therefore, the total number of mole of the gaseous products formed is 4.488 moles

C. Determination of the number of molecules of water vapor formed.

From Avogadro's hypothesis,

1 mole of water = 6.02×10²³ molecules

Therefore,

1.632 moles of water = 1.632 × 6.02×10²³

1.632 moles of water = 9.82×10²³ molecules

Thus, the number of molecules of water vapor formed is 9.82×10²³ molecules

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(A)

Density = Mass / Volume

So  

Mass = Density × Volume

Mole ratio of toluene : Oxygen is 1 : 9

(B)

1 mole of Toluene produces 7 moles of gas and 4 moles of Vapour

So the mole ratio is 1 : 11

(C)

1mole contains molecules

i think it is d. e because the next one is igneous.