Question

2 al(s) + 6 hcl(aq) → 2 alcl₃(aq) + 3 h₂(g)
according to the reaction represented above, about how many grams of aluminum (atomic mass 27 g) are
necessary to produce 0.50 mol of hydrogen gas at 25°c and 1.00 atm?
a 1.0 g
b 9.0 g
c 14 g
d 27 g

Explanation:

Step1: Determine the mole ratio of Al to H₂

From the balanced chemical equation \(2Al(s)+6HCl(aq)\to2AlCl_3(aq) + 3H_2(g)\), the mole ratio of \(Al\) to \(H_2\) is \(n(Al):n(H_2)=2:3\).

Step2: Calculate the moles of Al

Given \(n(H_2) = 0.50\space mol\). Using the mole ratio \(\frac{n(Al)}{n(H_2)}=\frac{2}{3}\), we can solve for \(n(Al)\). So \(n(Al)=\frac{2}{3}\times n(H_2)\). Substituting \(n(H_2) = 0.50\space mol\), we get \(n(Al)=\frac{2}{3}\times0.50\space mol=\frac{1}{3}\space mol\).

Step3: Calculate the mass of Al

The formula for mass \(m = n\times M\), where \(M\) (molar mass of \(Al\)) is \(27\space g/mol\) and \(n(Al)=\frac{1}{3}\space mol\). Then \(m(Al)=\frac{1}{3}\space mol\times27\space g/mol = 9.0\space g\).

Answer:

B. 9.0 g