Question #cac61

Mar 5, 2017

This appears to have no limiting reagent, as both reactants are present in the needed stoichiometric amount.

Explanation:

$C {H}_{5} N + C O C {l}_{2} \rightarrow 2 H C l + {C}_{2} {H}_{3} N O$

$31 g C {H}_{5} N = \frac{31 g}{31 \left(\frac{g}{\text{mol}}\right)} = 1$ mol

$98.91 g C O C {l}_{2} = \frac{98.91 g}{98.91 \left(\frac{g}{\text{mol}}\right)} = 1$ mol

So, we have one mole of each reactant. In the equation, the coefficient of each reactant is one, meaning that 1 mole of $C {H}_{5} N$ requires 1 mole $C O C {l}_{2}$ for complete reaction.

This is precisely what we are provided with in this problem, so it appears there is no limiting reagent here. Both chemicals run out at the same instant, and the yield is 100%.