$$ \ce{C3H8(g) + 4Cl2(g) -> C3H4Cl4(g) + 4HCl(g)} $$

A 6.0 mol sample of \(\ce{C3H8(g)}\) and a 20. mol sample of \(\ce{Cl2(g)}\) are placed in a previously evacuated vessel, where they react according to the equation above. After one of the reactants has been totally consumed, how many moles of \(\ce{HCl}\) have been produced?

Summary Submit Skip Question

Since neither reactant is given in excess, we need to find the limiting reactant. Since the ratio of \(\ce{C3H8}\) to \(\ce{Cl2}\) is 1:4,

$$ \frac{\text{mol }\ce{C3H8}}{\text{mol }\ce{Cl2}} = \frac{1}{4} = \frac{6}{x} $$ $$ x= 24 $$

Given 6 moles of \(\ce{C3H8}\), we need 24 moles of \(\ce{Cl2}\). Since we only have 20 moles of \(\ce{Cl2}\), \(\ce{Cl2}\) is the limiting reactant. We continue with stoichiometry to obtain the moles of \(\ce{HCl}\):

$$ \pu{20 mol }\ce{Cl2} \cdot \frac{\pu{4 mol }\ce{HCl}}{\pu{4 mol }\ce{Cl2}} $$ $$ = \pu{20 mol }\ce{HCl} $$