# Question b467f

Jul 9, 2017

Limiting reagent: ${\text{C"_2"H"_3"Br}}_{3}$

#### Explanation:

We're asked to find the limiting reactant in a chemical reaction given the masses of reactants.

Let's first write the balanced chemical equation for this reaction (look up what forms if you don't know for sure):

$4 {\text{C"_2"H"_3"Br"_3 (l) + 11"O"_2 (g) rarr 8"CO"_2 (g) + 6"H"_2"O" (g) + 6"Br}}_{2} \left(l\right)$

To find the limiting reactant, we convert the given masses of reactants to moles, then divide by the respective coefficient in front of it; the limiting reactant is the quantity whose number is lower:

$76.4 \cancel{\text{g C"_2"H"_3"Br"_3)((1color(white)(l)"mol C"_2"H"_3"Br"_3)/(266.76cancel("g C"_2"H"_3"Br"_3))) = 0.286/(4color(white)(l)"(coefficient)}}$

= color(red)(0.0716

$49.1 \cancel{\text{g O"_2)((1color(white)(l)"mol O"_2)/(32.00cancel("g O"_2))) = 1.53/(11color(white)(l)"(coefficient)}}$

= color(blue)(0.140

Since the number for ${\text{C"_2"H"_3"Br}}_{3}$ is lower, $\textcolor{red}{{\text{C"_2"H"_3"Br}}_{3}}$ sfcolor(red)("is the limiting reagent"#.