Question

Liquid hexane CH3CH24CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 5.2 g of hexane is mixed with 33.0 g of oxygen. Calculate the minimum mass of hexane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Answer:

The leftover is 15 g of oxygen rather than liquid hexane as the fuel limits the reaction.

Explanation:

Hello!

In this case, since the described combustion reaction is:

$C_6H_{14}(l)+\frac{19}{2} O_2(g)\rightarrow 6CO_2(g)+7H_2O(g)$

Thus, since 5.2 g of hexane (86.2 g/mol) is reacted with 33.0 g of oxygen (32.0 g/mol) we can compute the mass of hexane that was actually consumed via stoichiometry with oxygen (1:19/2 mole ratio):

$m_{C_6H_{14}}^{consumed \ by\ O_2}=33.0gO_2*\frac{1molO_2}{32.0gO_2}*\frac{1molC_6H_{14}}{\frac{19}{2}gO_2 } *\frac{86.2gC_6H_{14}}{1molC_6H_{14}} \\\\m_{C_6H_{14}}^{consumed \ by\ O_2}=9.36gC_6H_{14}$

It is proved then than the hexane won't have any leftover but oxygen does, as shown below:

$m_{O_2}^{consumed \ by\ C_6H_{14}}=5.2gC_6H_{14}*\frac{1molC_6H_{14}}{86.2gC_6H_{14}} *\frac{\frac{19}{2}molO_2 }{1molC_6H_{14}} *\frac{32.0gO_2}{1molO_2} \\\\m_{O_2}^{consumed \ by\ C_6H_{14}}=18g$

It means the leftover of oxygen is:

$m_{O_2}^{leftover}=33g-18g\\\\m_{O_2}^{leftover}=15g$

Regards!