Final answer:
To determine the amount of excess reactant remaining at the end of the reaction, compare the moles of each reactant present and their stoichiometric coefficients. Given 20.0g of HCl and 35.0g of Al, convert the masses to moles. From the balanced equation 2Al + 6HCl -> 2AlCl3 + 3H2, the stoichiometric ratio between Al and HCl is 1:3. HCl is the limiting reactant and Al is the excess reactant. Calculate the amount of excess reactant remaining by multiplying the number of moles of Al with the ratio of 2 moles AlCl3/6 moles HCl.
Step-by-step explanation:
To determine the amount of excess reactant remaining at the end of the reaction, we need to first identify the limiting reactant. This can be done by comparing the number of moles of each reactant present and their stoichiometric coefficients in the balanced chemical equation.
Given 20.0 g of HCl and 35.0 g of Al, we convert the masses to moles using the molar masses of HCl (36.46 g/mol) and Al (26.98 g/mol). This yields approximately 0.549 moles of HCl and 1.297 moles of Al.
From the balanced chemical equation 2Al + 6HCl -> 2AlCl3 + 3H2, we see that the stoichiometric ratio between Al and HCl is 1:3. Therefore, for every mole of Al, we would need 3 moles of HCl. Since we have less moles of HCl (0.549) compared to the stoichiometric ratio (3 * 1.297 = 3.891), HCl is the limiting reactant and Al is the excess reactant.
Finally, to determine how much of the excess reactant is remaining at the end of the reaction, we can use stoichiometry. For every mole of HCl reacted, 2 moles of AlCl3 are produced. Therefore, the amount of excess reactant remaining is calculated by multiplying the number of moles of Al by the ratio of 2 moles AlCl3/6 moles HCl.
Excess reactant remaining = 1.297 moles Al * (2 moles AlCl3/6 moles HCl) = 0.432 moles Al