Question

what is the limiting reactant and how much cr is produced when 24.0 g of al reacts with 40.0 g of cr₂o₃ according to the chemical equation below? the molar mass of cr₂o₃ is 152 g/mol.
2al + cr₂o₃ → al₂o₃ + 2cr
○ al is the limiting reactant and 46.2 g of cr is produced.
○ al is the limiting reactant and 27.4 g of cr is produced.
○ cr₂o₃ is the limiting reactant and 27.4 g of cr is produced.
○ al is the limiting reactant and 54.7 g of cr is produced.
○ cr₂o₃ is the limiting reactant and 46.2 g of cr is produced.

Explanation:

Step1: Calculate moles of reactants

Molar mass of \(Al = 27\space g/mol\). Moles of \(Al=\frac{24.0\space g}{27\space g/mol}\approx0.889\space mol\).
Moles of \(Cr_2O_3=\frac{40.0\space g}{152\space g/mol}\approx0.263\space mol\).

Step2: Determine moles of \(Cr\) from each reactant

From the balanced equation \(2Al + Cr_2O_3
ightarrow Al_2O_3 + 2Cr\), mole ratio of \(Al:Cr = 2:2 = 1:1\). Moles of \(Cr\) from \(Al=0.889\space mol\).
Mole ratio of \(Cr_2O_3:Cr = 1:2\). Moles of \(Cr\) from \(Cr_2O_3=0.263\times2 = 0.526\space mol\).
Since \(Cr_2O_3\) gives less moles of \(Cr\), \(Cr_2O_3\) is the limiting reactant.

Step3: Calculate mass of \(Cr\)

Molar mass of \(Cr = 52\space g/mol\). Mass of \(Cr=0.526\space mol\times52\space g/mol\approx27.4\space g\)

Answer:

\(Cr_2O_3\) is the limiting reactant and \(27.4\space g\) of \(Cr\) is produced. So the answer is: \(Cr_2O_3\) is the limiting reactant and \(27.4\space g\) of \(Cr\) is produced.