Question

(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g of C8H18 (a typical component of gasoline)? Page 253 (b) For part (a), the source of O2 is air, which is 78% N2, 21% O2, and 1.0% Ar by volume. Assuming all the O2 reacts, but no N2 or Ar does, what is the total volume (in L) of the engine’s gaseous exhaust?

Answer 1

Answer:

Part A

 The volume of the gaseous product  is  $V = 787L$

Part B

The volume of the the engine’s gaseous exhaust is  $V_e = 2178 \ L$

Explanation:

Part A

From the question we are told that

    The temperature is  $T = 350^oC = 350 +273 =623K$

     The pressure is  $P = 735 \ torr = \frac{735}{760} = 0.967\ atm$

     The of  $C_8 H_{18} = 100.0g$

The chemical equation for this combustion is

               $2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}$

 The number of moles of  $C_8 H_{18}$ that reacted is mathematically represented as

               $n = \frac{mass \ of \ C_8H_{18} }{Molar \ mass \ of C_8H_{18} }$

The molar mass of  $C_8 H_{18}$ is constant value which is

                  $M = 114.23 \ g/mole$  

So          $n = \frac{100 }{114.23} }$

             $n = 0.8754 \ moles$

The gaseous product in the reaction is $CO_2_{(g)}$ and water vapour

Now from the reaction

    2 moles of $C_8 H_{18}$  will react with 25 moles of $O_2$ to give (16 + 18) moles of $CO_2_{(g)}$ and  $H_2 O_{(g)}$

So

    1 mole of $C_8 H_{18}$ will  react with 12.5 moles of  $O_2$ to give 17 moles of $CO_2_{(g)}$ and  $H_2 O_{(g)}$

This implies that

    0.8754 moles of $C_8 H_{18}$ will react with (12.5 * 0.8754 ) moles of $O_2$ to give  (17 * 0.8754) of $CO_2_{(g)}$ and  $H_2 O_{(g)}$

So the no of moles of gaseous product is

         $N_g = 17 * 0.8754$

         $N_g = 14.88 \ moles$

From the ideal gas law

       $PV = N_gRT$

making V the subject

        $V = \frac{N_gRT}{P}$

Where R is the gas constant with a value $R = 0.08206 \ L\cdot atm /K \cdot mole$

Substituting values

          $V = \frac{14.88* 0.08206 *623}{0.967}$

          $V = 787L$

Part B

From the reaction the number of moles of oxygen that reacted is

         $N_o = 0.8754 * 12.5$

         $N_o = 10.94 \ moles$

The volume is

      $V_o = \frac{10.94 * 0.08206 *623}{0.967}$

      $V_o = 579 \ L$

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

         $V_e = V_o * \frac{0.79}{0.21}$

Substituting values

       $V_e = 579 * \frac{0.79}{0.21}$

       $V_e = 2178 \ L$