Respuesta :
Answer:
(A)
Density = Mass / Volume
So
Mass = Density × Volume
[tex]= 0.867 g/mL \times 43.3mL = 37.5411 g Toluene[/tex]
[tex]1C_6 H_5 CH_3 + 9 O_2 > 7 CO_2 + 4 H_2 O[/tex]
Mole ratio of toluene : Oxygen is 1 : 9
[tex]$37.5411 g \text { Toluene } \times \frac{1 \text {mol} \text {toluene}}{92 g \text { toluene}} \times \frac{9 {mol} O_{2}}{1 \text {mol} \text { toluene }} \times \frac{32 g O_{2}}{1 {mol} O_{2}}=117 g O_{2}(\text {Answer})$[/tex]
(B)
1 mole of Toluene produces 7 moles of [tex]CO_2[/tex] gas and 4 moles of [tex]H_2 O[/tex] Vapour
So the mole ratio is 1 : 11
[tex]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) } $[/tex]
(C)
1mole contains [tex]6.022\times10^{23}[/tex] molecules
[tex]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) } $[/tex]
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 g
- Next, we shall write the balanced equation for the reaction.
C₆H₅CH₃ + 9O₂ —> 7CO₂ + 4H₂O
- Next, 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
Learn more: https://brainly.com/question/15331762
