Question

Question #b467f

Answer 1

Limiting reagent: `"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"C"_2"H"_3"Br"_3 (l) + 11"O"_2 (g) rarr 8"CO"_2 (g) + 6"H"_2"O" (g) + 6"Br"_2 (l)`

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:

`76.4cancel("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`

`49.1cancel("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 `"C"_2"H"_3"Br"_3` is lower, `color(red)("C"_2"H"_3"Br"_3)` `sfcolor(red)("is the limiting reagent"`.