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solmaris [256]
3 years ago
8

Consider the combustion of octane (C8H18), a primary component of gasoline.

Chemistry
1 answer:
kicyunya [14]3 years ago
4 0

Answer:

a. 2562 L

b. 4082 L

c. 413,6 kJ

d. 0,115 kWh

e. 1,3 cents

f. 0,34 %

Explanation:

a. To calculate the volume we should obtain the moles of Octane to know the reactant moles and produced moles. Then, with ideal gas law obtain the change in volume:

3,784 L ≡ 3784 mL octane × (0,703 g / 1 mL) = <em>2660 g Octane</em>

<em>                                                    density</em>

2660 g octane × ( 1 mol / 114,23 g octane) = <em>23,29 mol octane</em>

                              <em>molar mass of C₈H₁₈</em>

With the combustion reaction of octane we can know how many moles are produced from 23,29 mol of octane, thus, in (1) :

2 C₈H₁₈ (l) + 25 O₂(g) ---> 18 H₂O(g) + 16 CO₂ (g) <em>(1)</em>

23,29 mol octane × ( 25 mol O₂ / 2 mol octane) = <em>291,1 mol O₂ -reactant moles-</em>

23,29 mol octane × ( 18 + 16 produced mol  / 2 mol octane) = 395,9 moles produced

Ideal gas formula says:

V = nRT/ P

Where:

n = Δmoles number (produced-reactant) → 104,8 moles

R = Ideal gas constant → 0,082 atm·L/mol·K

T = Temperature → 25°C, 298,15 K

P = Pressure → 1 atm

Thus, replacing in the equation:

ΔV = 2562 L

b. To calculate the gas volume we should use the same values of ideal gas formula just changing the temperature value for 475 K -Because the produced moles of gas and presure are the same and R is constant.

Thus, the volume of produced gases is:

ΔV = 4082 L

c. The work, w, is equal to -pressure times Δ Volume:

w = - P×ΔV

The pressure is 1 atm and ΔV in the system is 4082 L

So, w = 4082 atm·L (101,325 J / 1 atm·L) = 413,6 kJ

d. As kJ is equal to kWs, 413,6 kJ ≡ 413,6 kWs × ( 1 hour / 3600 s) =

0,115 kWh

e. In Seattle 1kWh cost 11,35 cents. So, 0,115 kWh cost:

0,115 kWh × (11,35 cents/ 1kWh) = 1,3 cents

f. The energy calculated in part C, <em>413,6 kJ</em> is due to the work done by the system in gas expansion but total of heat produced in (1) are 1,2 ×10⁵ kJ. Thus, the proportion of work in gas expansion in total energy in combustion of octane is:

413,6 kJ / 1,2×10 ⁵ kJ × 100 = 0,34 %

I hope it helps!

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Answer:

34.7mL

Explanation:

First we have to convert our grams of Zinc to moles of zinc so we can relate that number to our chemical equation.

So: 6.25g Zn x (1 mol / 65.39 g) = 0.0956 mol Zn

All that was done above was multiplying the grams of zinc by the reciprocal of zincs molar mass so our units would cancel and leave us with moles of zinc.

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Karo-lina-s [1.5K]

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Explanation:

Hello, for each statement we state:

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