Answer :
To determine which reactant is in excess and by how many grams, we need to compare the stoichiometry and molar masses of Potassium Phosphate (K3PO4) and Calcium Nitrate (Ca(NO3)2). excess by 35.4 grams.
Given the masses of each reactant, we can calculate the number of moles of each substance and determine the stoichiometric ratio. The reactant that has a higher ratio is the limiting reactant, while the one with a lower ratio is in excess.
The difference in the masses of the two reactants will indicate by how many grams the excess reactant exceeds the stoichiometric requirement.
The molar mass of Potassium Phosphate (K3PO4) can be calculated as follows:
(3 * atomic mass of potassium) + atomic mass of phosphorus + (4 * atomic mass of oxygen) = (3 * 39.1 g/mol) + 31.0 g/mol + (4 * 16.0 g/mol) = 212.3 g/mol
The molar mass of Calcium Nitrate (Ca(NO3)2) is calculated as:
atomic mass of calcium + (2 * atomic mass of nitrogen) + (6 * atomic mass of oxygen) = 40.1 g/mol + (2 * 14.0 g/mol) + (6 * 16.0 g/mol) = 164.1 g/mol
Next, we calculate the number of moles for each reactant using the given masses:
moles of Potassium Phosphate = 83.6 g / 212.3 g/mol = 0.393 mol
moles of Calcium Nitrate = 98.5 g / 164.1 g/mol = 0.600 mol
The stoichiometric ratio between Potassium Phosphate and Calcium Nitrate is 1:1, indicating that one mole of each reactant is required to react completely. Since the moles of Calcium Nitrate (0.600 mol) are greater than the moles of Potassium Phosphate (0.393 mol), Calcium Nitrate is in excess.
To determine the excess amount of Calcium Nitrate, we calculate the difference in mass:
mass of excess Calcium Nitrate = (moles of Calcium Nitrate - moles of Potassium Phosphate) * molar mass of Calcium Nitrate
mass of excess Calcium Nitrate = (0.600 mol - 0.393 mol) * 164.1 g/mol = 35.4 g
Therefore, Calcium Nitrate is in excess by 35.4 grams.
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