Question

2. Consider the reaction: 2C6H4Cl₂ + 130₂ 12CO₂ + 2H₂O + 4HC14a.) If you had 18.5 mol of C6H4Cl2, how many mol of CO2 would you produce?b.) If you had 150.6 mol of O2, how many mol of CO2 would you produce?c.) If you had 18.5 mol of C6H4Cl2 and 150.6 mol of O2, which reactant would be limiting?d.) What is the theoretical yield of CO2, in moles?

Answer 1

Answer and Explanation

Given the equation

$2C_6H_4Cl_2+13O_2\rightarrow12CO_2+2H_2O+4HCl$

(a)

$\begin{gathered} 18.5\text{ mol C}_6H_4Cl_2\text{ x }\frac{12\text{ mol CO}_2}{2\text{ mol C}_2H_4Cl_2} \\ \\ =\text{ 111 mol CO}_2 \end{gathered}$

(b)

$\begin{gathered} 150.6\text{ mol O}_2\text{ x }\frac{12\text{ mol CO}_2}{13\text{ mol O}_2} \\ \\ =\text{ 139.01 mol CO}_2 \end{gathered}$

(c) C6H4Cl₂ is the limiting reactant because it will produce less amount of CO2 moles.

(d) The theoretical yield of CO2:

m = n x M where m is the mass (theoretical yield), n is the moles and M is the molar mass of CO2

m = 111 mol x 44,01 g/mol

m = 4885.11 g