Question

find the mass of alcl₃ that is produced when 25.0 grams of al₂o₃ reacts with hcl according to the following equation. al₂o₃ + 6hcl → 2alcl₃ + 3h₂o 65.3 g 32.6 g 16.3 g 155 g 72.9 g

Explanation:

Step1: Calculate the molar mass of \(Al_2O_3\)

The molar mass of \(Al\) is \(27\space g/mol\) and of \(O\) is \(16\space g/mol\).
For \(Al_2O_3\), \(M = 2\times27+3\times16= 54 + 48=102\space g/mol\)

Step2: Find the moles of \(Al_2O_3\)

Using the formula \(n=\frac{m}{M}\), where \(m = 25.0\space g\) and \(M = 102\space g/mol\)
\(n=\frac{25.0}{102}\space mol\approx0.245\space mol\)

Step3: Use the stoichiometry from the balanced equation

From \(Al_2O_3+6HCl
ightarrow2AlCl_3 + 3H_2O\), the mole ratio of \(Al_2O_3\) to \(AlCl_3\) is \(1:2\)
So moles of \(AlCl_3=2\times\)moles of \(Al_2O_3\)
Moles of \(AlCl_3 = 2\times0.245 = 0.49\space mol\)

Step4: Calculate the molar mass of \(AlCl_3\)

The molar mass of \(Al\) is \(27\space g/mol\) and of \(Cl\) is \(35.5\space g/mol\)
For \(AlCl_3\), \(M=27 + 3\times35.5=27+106.5 = 133.5\space g/mol\)

Step5: Calculate the mass of \(AlCl_3\)

Using \(m=n\times M\), where \(n = 0.49\space mol\) and \(M = 133.5\space g/mol\)
\(m=0.49\times133.5\space g\approx65.4\space g\) (close to \(65.3\space g\) due to rounding differences in intermediate steps)

Answer:

A. \(65.3\space g\)