Step-by-step explanation:
To determine which chemical is in excess, we need to compare the number of moles of each reactant used to the stoichiometric ratio of the reaction. The balanced equation tells us that 1 mole of Ca(OH)2 reacts with 2 moles of HCl. Therefore, for every mole of Ca(OH)2 used, we need 2 moles of HCl to react completely.
Using the given amounts, we have:
Moles of Ca(OH)2 = 3.0 mol
Moles of HCl = 10.0 mol
To determine which chemical is in excess, we can calculate how many moles of HCl are needed to react completely with 3.0 mol of Ca(OH)2:
2 moles HCl / 1 mole Ca(OH)2 * 3.0 mol Ca(OH)2 = 6.0 mol HCl needed
Since we have 10.0 mol of HCl, which is more than the 6.0 mol needed to react completely with 3.0 mol of Ca(OH)2, HCl is in excess.
b. The excess is the difference between the moles of the excess reactant and the moles needed to react completely with the limiting reactant. In this case, we have:
Moles of HCl used = 6.0 mol (needed to react with 3.0 mol Ca(OH)2)
Moles of HCl available = 10.0 mol
Excess HCl = 10.0 mol - 6.0 mol = 4.0 mol
Therefore, the excess of HCl is 4.0 mol.
c. The stoichiometry of the reaction tells us that for every mole of Ca(OH)2 used, 2 moles of water are produced. Since we have used 3.0 moles of Ca(OH)2, theoretically, we can produce:
2 moles H2O / 1 mole Ca(OH)2 * 3.0 mol Ca(OH)2 = 6.0 moles of H2O
Therefore, theoretically, 6.0 moles of H2O will be produced.