Question

page 2:
what is the limiting reagent if 76.4 grams of c2h3br3 were reacted with 49.1 grams of o2?
4c2h3br3 + 11o2 → 8co2 + 6h2o + 6br2
c2h3br3
o2
co2
br2
h2o

Explanation:

Step1: Calculate molar mass

Molar mass of \(C_2H_3Br_3\):
\(M_{C_2H_3Br_3}=(2\times12.01)+(3\times1.01)+(3\times79.90)=24.02 + 3.03+239.7=266.75\space g/mol\)
Molar mass of \(O_2\): \(M_{O_2}=2\times16.00 = 32.00\space g/mol\)

Step2: Calculate moles of reactants

Moles of \(C_2H_3Br_3\): \(n_{C_2H_3Br_3}=\frac{m}{M}=\frac{76.4\space g}{266.75\space g/mol}\approx0.286\space mol\)
Moles of \(O_2\): \(n_{O_2}=\frac{m}{M}=\frac{49.1\space g}{32.00\space g/mol}\approx1.534\space mol\)

Step3: Use stoichiometry

From the balanced equation \(4C_2H_3Br_3+11O_2
ightarrow8CO_2 + 6H_2O+6Br_2\), the mole ratio of \(C_2H_3Br_3\) to \(O_2\) is \(\frac{4}{11}\)
If all \(C_2H_3Br_3\) reacts, moles of \(O_2\) needed: \(n_{O_2,needed}=0.286\space mol\times\frac{11}{4}=0.788\space mol\)
Since \(0.788\space mol<1.534\space mol\) (moles of \(O_2\) available), \(C_2H_3Br_3\) is the limiting reactant.

Answer:

\(C_2H_3Br_3\)