6.12.67 moles of aluminum react with 33.54 grams of hydrochloric acid Determine the moles of hydrogen and identify the limiting and excess reactants. Balanced Equation: __

## Step 1<br />The balanced chemical equation for the reaction between aluminum and hydrochloric acid is:<br />### \(2Al + 6HCl \rightarrow 2AlCl3 + 3H2\)<br />This equation tells us that 2 moles of aluminum react with 6 moles of hydrochloric acid to produce 3 moles of hydrogen.<br /><br />## Step 2<br />We are given that 12.67 moles of aluminum are present. According to the balanced equation, 2 moles of aluminum require 6 moles of hydrochloric acid. Therefore, 12.67 moles of aluminum would require \(12.67 \times \frac{6}{2} = 37.01\) moles of hydrochloric acid.<br /><br />## Step 3<br />However, we only have 33.54 grams of hydrochloric acid, which is equivalent to \(33.54 \div 36.5 = 0.92\) moles (since the molar mass of hydrochloric acid is 36.5 g/mol).<br /><br />## Step 4<br />Since we need 37.01 moles of hydrochloric acid to react with 12.67 moles of aluminum, but we only have 0.92 moles of hydrochloric acid, we have less than the required amount. Therefore, hydrochloric acid is the limiting reactant.<br /><br />## Step 5<br />Since we have more than the required amount of aluminum, it is the excess reactant.<br /><br />## Step 6<br />To find the moles of hydrogen produced, we use the stoichiometry of the reaction. According to the balanced equation, 6 moles of hydrochloric acid produce 3 moles of hydrogen. Therefore, 0.92 moles of hydrochloric acid will produce \(0.92 \times \frac{3}{6} = 0.46\) moles of hydrogen.