Chemistry, 01.10.2019 01:30 onlineK

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?

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Answer from: seby951

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 of toluene =?

Mass = Density × Volume

Mass of toluene = 0.867 × 43.3

Mass of toluene = 37.5411 gNext, we shall write the balanced equation for the reaction. C₆H₅CH₃ + 9O₂ —> 7CO₂ + 4H₂ONext, we shall determine the masses of C₆H₅CH₃ and O₂ that reacted from the balanced equation

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

We'll begin by calculating the number of mole of in 37.5411 g of C₆H₅CH₃

Molar mass of C₆H₅CH₃ = 92 g/mol

Mass of C₆H₅CH₃ = 37.5411 g

Mole of C₆H₅CH₃ =?

Mole = mass / molar mass

Mole of C₆H₅CH₃ = 37.5411 / 92

Mole of C₆H₅CH₃ = 0.408 mole Next, we shall determine the mole of CO₂ produced.

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

Next, we shall determine the mole of H₂O produced.

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

Finally, we shall determine the total moles of the gaseous products

Mole of CO₂ produced = 2.856 moles

Mole of H₂O produced = 1.632 moles

Total mole = 2.856 + 1.632

Total mole = 4.488 moles

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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Answer from: vanitycarraway2000

(A)

Density = Mass / Volume

Mass = Density × Volume

$= 0.867 g/mL \times 43.3mL = 37.5411 g Toluene$

1C_6 H_5 CH_3 + 9 O_2 7 CO_2 + 4 H_2 O

Mole ratio of toluene : Oxygen is 1 : 9

$37.5411 g \text { Toluene } \times \frac{1 \text {mol} \text {toluene}}{92 g \text { toluene}} \times \frac{9 {mol} O_{2}}{1 \text {mol} \text { toluene }} \times \frac{32 g O_{2}}{1 {mol} O_{2}}=117 g O_{2}(\text {Answer})$

(B)

1 mole of Toluene produces 7 moles of CO_2 gas and 4 moles of H_2 O Vapour

So the mole ratio is 1 : 11

$37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene }} \times \frac{11 \mathrm{mol} \text { gas }}{1 \text { mol toluene }} $$\\\\=4.49 \text { mol gaseous products (Answer) } $$

(C)

1mole contains $6.022\times10^{23}$ molecules

$37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene}} \times \frac{4 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \text { toluene }} \times \frac{6.022 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}} $\\\\$=9.82 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O} \text { (Answer) } $$

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One of the compounds used to increase the octane rating of gasoline is toluene (pictured). suppose 4...

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