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4. If 72.0 grams of aluminum are reacted with 252 grams of hydrochloric acid, what is the limiting reactant?2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)AluminumHydrochloric acidAluminum chlorideHydrogen gas

Question

  1. If 72.0 grams of aluminum are reacted with 252 grams of hydrochloric acid, what is the limiting reactant?2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)AluminumHydrochloric acidAluminum chlorideHydrogen gas
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Solution

To determine the limiting reactant, we first need to calculate the number of moles of each reactant.

  1. Calculate the molar mass of Aluminum (Al) and Hydrochloric acid (HCl). The molar mass of Al is approximately 26.98 g/mol and the molar mass of HCl is approximately 36.46 g/mol.

  2. Calculate the number of moles for each reactant.

    For Aluminum: Number of moles = mass / molar mass = 72.0 g / 26.98 g/mol = 2.67 moles

    For Hydrochloric acid: Number of moles = mass / molar mass = 252 g / 36.46 g/mol = 6.91 moles

  3. Compare the mole ratio of the reactants with the coefficients in the balanced chemical equation.

    The balanced chemical equation is: 2Al + 6HCl → 2AlCl3 + 3H2

    The ratio of Al to HCl in the balanced equation is 2:6 or 1:3.

    The ratio of Al to HCl in our reaction is 2.67 moles : 6.91 moles or approximately 1:2.6

  4. The reactant that is not in excess is the limiting reactant.

    Since the ratio of Al to HCl in our reaction (1:2.6) is less than the ratio in the balanced equation (1:3), we have less HCl than required. Therefore, Hydrochloric acid (HCl) is the limiting reactant.

This problem has been solved

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