Question

now lets say we have:
1.64 moles of al
2.42 moles of hbr
using the equation above to answer the the following questions.
need help determining the limiting reactant?
want to know how to calculate the amount of albr₃ formed?
how much of the excess reactant is left over?
what is the limiting reactant?
given the limiting reactant, how much albr₃ would form?
moles
how much of the excess reactant would be left over?
moles

Explanation:

Step1: Write the balanced chemical equation

The reaction between \(Al\) and \(HBr\) is \(2Al + 6HBr=2AlBr_{3}+3H_{2}\).

Step2: Calculate the moles of \(AlBr_{3}\) if \(Al\) is the limiting reactant

From the balanced equation, the mole ratio of \(Al\) to \(AlBr_{3}\) is \(1:1\). If \(n(Al) = 1.64\space mol\), then \(n(AlBr_{3})\) (if \(Al\) is limiting) is \(1.64\space mol\)

Step3: Calculate the moles of \(AlBr_{3}\) if \(HBr\) is the limiting reactant

The mole ratio of \(HBr\) to \(AlBr_{3}\) is \(3:1\). Given \(n(HBr)=2.42\space mol\), then \(n(AlBr_{3})=\frac{2.42}{3}\space mol\approx0.81\space mol\)

Step4: Determine the limiting reactant

Since \(0.81\space mol<1.64\space mol\), \(HBr\) is the limiting reactant.

Step5: Calculate the moles of \(AlBr_{3}\) formed

As \(HBr\) is the limiting reactant, \(n(AlBr_{3})=\frac{2.42}{3}\space mol\approx0.81\space mol\)

Step6: Calculate the moles of excess reactant (\(Al\)) left over

Moles of \(Al\) that react: From the mole ratio \(2Al:6HBr\), moles of \(Al\) that react with \(2.42\space mol\space HBr\) is \(\frac{2}{6}\times2.42\space mol = 0.81\space mol\)
Moles of \(Al\) left over: \(n(Al)_{initial}-n(Al)_{reacted}=1.64 - 0.81= 0.83\space mol\)

Answer:

Limiting reactant: \(HBr\)
Moles of \(AlBr_{3}\) formed: \(0.81\space mol\)
Moles of excess reactant (\(Al\)) left over: \(0.83\space mol\)