Superphosphate, or sometimes called triple phosphate, is a commonly used water-soluble
fertilizer. It is produced by the reaction:
πΆπ3 (ππ4)2 + π»2ππ4 β πΆπ(π»2ππ4)2 + πΆπππ4 If we mix 200.0 g of πΆπ3(ππ4)2 with 133.5 g of π»2ππ4,
how much πΆπ(π»2ππ4)2 (in grams)
can be produced?
To find out how much Ca(H2PO4)2 can be produced, we need to determine the limiting reactant first.
Step 1: Calculate the number of moles for each reactant.
Given:
Mass of Ca3(PO4)2 = 200.0 g
Mass of H2SO4 = 133.5 g
Molar mass of Ca3(PO4)2 = 310.18 g/mol
Molar mass of H2SO4 = 98.09 g/mol
Number of moles of Ca3(PO4)2 = mass / molar mass = 200.0 g / 310.18 g/mol
Number of moles of Ca3(PO4)2 = 0.644 mol
Number of moles of H2SO4 = mass / molar mass = 133.5 g / 98.09 g/mol
Number of moles of H2SO4 = 1.36 mol
Step 2: Determine the limiting reactant.
To determine the limiting reactant, we need to compare the stoichiometric ratio between Ca3(PO4)2 and H2SO4.
From the balanced equation:
1 mole of Ca3(PO4)2 produces 1 mole of Ca(H2PO4)2
1 mole of H2SO4 produces 1 mole of Ca(H2PO4)2
Based on these ratios, the mole ratio between Ca3(PO4)2 and Ca(H2PO4)2 is 1:1, irrespective of H2SO4.
Step 3: Calculate the maximum possible amount of Ca(H2PO4)2 that can be produced.
Since the mole ratio of Ca3(PO4)2 and Ca(H2PO4)2 is 1:1, the maximum number of moles of Ca(H2PO4)2 that can be produced is equal to the number of moles of Ca3(PO4)2.
Number of moles of Ca(H2PO4)2 = 0.644 mol
Step 4: Calculate the mass of Ca(H2PO4)2.
Mass of Ca(H2PO4)2 = number of moles Γ molar mass
Mass of Ca(H2PO4)2 = 0.644 mol Γ (120.09 g/mol) [Molar mass of Ca(H2PO4)2]
Mass of Ca(H2PO4)2 = 77.68 g
Therefore, the maximum amount of Ca(H2PO4)2 that can be produced is 77.68 grams.
To find out how much Ca(H2PO4)2 can be produced, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed in a chemical reaction, thereby determining the maximum amount of product that can be formed.
To determine the limiting reactant, we need to compare the number of moles of each reactant. First, let's calculate the number of moles for each reactant:
Mass of Ca3(PO4)2 = 200.0 g
Molar mass of Ca3(PO4)2 = 310.18 g/mol
Number of moles of Ca3(PO4)2 = Mass / Molar mass
= 200.0 g / 310.18 g/mol
Mass of H2SO4 = 133.5 g
Molar mass of H2SO4 = 98.09 g/mol
Number of moles of H2SO4 = Mass / Molar mass
= 133.5 g / 98.09 g/mol
Now, let's compare the moles of each reactant to determine the limiting reactant.
The balanced equation tells us that 1 mole of Ca3(PO4)2 reacts with 2 moles of H2SO4 to produce 1 mole of Ca(H2PO4)2. Therefore, using the moles of each reactant, we can determine the moles of Ca(H2PO4)2 that can be produced.
From the balanced equation, we can see that the ratio of Ca3(PO4)2 to Ca(H2PO4)2 is 1:1. So, the number of moles of Ca3(PO4)2 is equal to the number of moles of Ca(H2PO4)2.
Since the moles of both reactants are equal if they react completely, we can conclude that the limiting reactant is the one with fewer moles. Therefore, we need to compare the moles of Ca3(PO4)2 and H2SO4.
Number of moles of Ca3(PO4)2 = 200.0 g / 310.18 g/mol
β 0.644 mol
Number of moles of H2SO4 = 133.5 g / 98.09 g/mol
β 1.361 mol
Since we have less Ca3(PO4)2 (0.644 mol) compared to the moles of H2SO4 (1.361 mol), Ca3(PO4)2 is the limiting reactant.
Now, we can use the balanced equation to determine the moles of Ca(H2PO4)2 that can be produced.
From the balanced equation, we know that 1 mole of Ca3(PO4)2 reacts to produce 1 mole of Ca(H2PO4)2. Therefore, the moles of Ca(H2PO4)2 produced will be the same as the moles of Ca3(PO4)2.
Number of moles of Ca(H2PO4)2 = 0.644 mol
To convert the moles to grams, we need to multiply by the molar mass of Ca(H2PO4)2:
Molar mass of Ca(H2PO4)2 = 234.04 g/mol
Mass of Ca(H2PO4)2 = Number of moles Γ Molar mass
= 0.644 mol Γ 234.04 g/mol
Therefore, approximately 150.29 grams of Ca(H2PO4)2 can be produced when 200.0 g of Ca3(PO4)2 is reacted with 133.5 g of H2SO4.