Answer :
Answer
26.82 grams Al
Explanation
Given equation:
[tex]2Al\mleft(s\mright)+6HCl\mleft(aq\mright)\to2AlCl_3\mleft(aq\mright)+3H_2\mleft(g\mright)[/tex]Given number of moles of HCl = 2.98 mol
What to find:
The mass of aluminum needed to completely react with 2.98 mol of hydrochloric acid.
Step-by-step solution:
The first step is to determine the mole of Al that would be needed to react completely with 2.98 moles of HCl by comparing the mole with the balanced equation.
From the balanced equation; 6 mol HCl reacts with 2 mol Al,
Therefore, 2.98 mol HCl will react with
[tex]\frac{2.98\text{ mol HCl }\times2\text{ mol Al}}{6\text{ mo HCl}}=0.9933\text{ mol Al}[/tex]The number of moles of Al that would react completely with 2.98 mol HCl = 0.9933 mol.
The last step is to determine the mass of Al present in 0.9933 mol Al by using the mole formula below:
[tex]\begin{gathered} \text{Number of moles }=\frac{Mass}{Molar\text{ mass}} \\ So, \\ \text{Mass }=Number\text{ of moles }\times Molar\text{ mass} \end{gathered}[/tex]From the Periodic Table; Molar mass of Al = 27.0 g/mol
Therefore, Mass of Al = 0.9933 mol x 27.0 g/mol = 26.82 grams
