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dexar [7]
3 years ago
14

Here is the combustion reaction for octane (c8h18), which is a primary component of gasoline. how many moles of co2 are emitted

into the atmosphere when 15.1 g of c8h18 is burned?
Chemistry
1 answer:
Ivenika [448]3 years ago
3 0
<span>Combustion reaction for octane is as follows.
2C</span>₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

Moles (mol) = mass (g) / molar mass (g/mol)

mass of octane = 15.1 g
molar mass of octane = <span>114.23 g/mol
Hence, moles of octane = 15. 1 g / </span><span>114.23 g/mol
                                       = 0.132 mol

The stoichiometric ratio between </span>C₈H₁₈ and CO₂ is 1 : 8.
Hence,
    moles of CO₂ produced = moles of octane burnt x 8
                                          = 0.132 mol x 8
                                          = 1.056 mol

Hence, 15.1 g of octane produce 1.056 moles of CO₂.
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Answer:

Percentage dissociated = 0.41%

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C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}

The ICE table is then shown as:

                               C_3H_6ClCO_2H_{(aq)}  \ \ \ \ \to  \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ +  \ \ \ \ H^+_{(aq)}

Initial   (M)                     1.8                                       0                               0

Change  (M)                   - x                                     + x                           + x

Equilibrium   (M)            (1.8 -x)                                  x                              x

K_a  = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}

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K_a = 3.02*10^{-5}

3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}

Since the value for K_a is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

3.02*10^{-5} *(1.8) = {(x)(x)}

5.436*10^{-5}= {(x^2)

x = \sqrt{5.436*10^{-5}}

x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M

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Percentage dissociated = \frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100

Percentage dissociated = \frac{7.3729*10^{-3}}{1.8 }*100

Percentage dissociated = 0.4096

Percentage dissociated = 0.41%     (to two significant digits)

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