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:
[tex]C_6H_{14}(l)+\frac{19}{2} O_2(g)\rightarrow 6CO_2(g)+7H_2O(g)[/tex]
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):
[tex]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}[/tex]
It is proved then than the hexane won't have any leftover but oxygen does, as shown below:
[tex]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[/tex]
It means the leftover of oxygen is:
[tex]m_{O_2}^{leftover}=33g-18g\\\\m_{O_2}^{leftover}=15g[/tex]
Regards!
