Determine the starting mass of each reactant if 46.3 of K3PO4 is produced and 92.8 of H3PO4 remains unreacted.
3KOH + H3PO4 ----> K3PO4 + 3H2O
To determine the starting mass of each reactant, we need to use the mole ratio from the balanced chemical equation.
The balanced equation is:
3KOH + H3PO4 → K3PO4 + 3H2O
From the equation, we can see that the ratio of H3PO4 to K3PO4 is 1:1.
First, we need to convert the given mass of K3PO4 produced (46.3 g) into moles. To do this, we divide the mass by the molar mass of K3PO4:
Molar mass of K3PO4 = (3 × molar mass of K) + molar mass of P + (4 × molar mass of O)
Molar mass of K = 39.1 g/mol
Molar mass of P = 30.97 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of K3PO4 = (3 × 39.1 g/mol) + 30.97 g/mol + (4 × 16.00 g/mol)
= 123.3 g/mol + 30.97 g/mol + 64.00 g/mol
= 218.27 g/mol
Now, we can calculate the number of moles of K3PO4 produced:
Number of moles = mass / molar mass
Number of moles of K3PO4 = 46.3 g / 218.27 g/mol
≈ 0.212 moles
Since the ratio of H3PO4 to K3PO4 is 1:1, the number of moles of H3PO4 initially present is also approximately 0.212 moles.
Now we need to find the starting mass of H3PO4. To do this, we multiply the number of moles by its molar mass:
Mass of H3PO4 = number of moles × molar mass
≈ 0.212 moles × (3 × molar mass of H + molar mass of P + 4 × molar mass of O)
≈ 0.212 mol × (3 × (1.01 g/mol) + 30.97 g/mol + 4 × (16.00 g/mol))
≈ 0.212 mol × (3.03 g/mol + 30.97 g/mol + 64.00 g/mol)
≈ 0.212 mol × (98.00 g/mol)
≈ 21 g
Therefore, the starting mass of H3PO4 is approximately 21 grams.
To determine the starting mass of each reactant, we need to use stoichiometry. Stoichiometry allows us to relate the masses of reactants with the mass of the product.
The balanced chemical equation for the reaction is:
3KOH + H3PO4 -> K3PO4 + 3H2O
From this equation, we can see that the ratio between KOH and K3PO4 is 3:1 and the ratio between H3PO4 and K3PO4 is 1:1.
Let's first calculate the mass of K3PO4 produced:
Given: Mass of K3PO4 = 46.3 g
Since the molar mass of K3PO4 is 212.27 g/mol, we can use the equation:
moles of K3PO4 = mass of K3PO4 / molar mass of K3PO4
moles of K3PO4 = 46.3 g / 212.27 g/mol = 0.2182 mol
Since the ratio between H3PO4 and K3PO4 is 1:1, the number of moles of H3PO4 remaining unreacted is also 0.2182 mol.
Now, let's calculate the mass of H3PO4:
Given: Mass of H3PO4 remaining = 92.8 g
Again, we can use the equation:
moles of H3PO4 = mass of H3PO4 / molar mass of H3PO4
moles of H3PO4 = 92.8 g / 98.00 g/mol = 0.947 mol
Since the ratio between H3PO4 and KOH is 1:3, the number of moles of KOH used is three times the number of moles of H3PO4:
moles of KOH = 3 * moles of H3PO4 = 3 * 0.2182 mol = 0.6546 mol
Finally, let's calculate the mass of KOH:
Given: Molar mass of KOH = 56.11 g/mol
mass of KOH = moles of KOH * molar mass of KOH
mass of KOH = 0.6546 mol * 56.11 g/mol = 36.7 g
Therefore, the starting mass of each reactant is:
Starting mass of KOH = 36.7 g
Starting mass of H3PO4 = 92.8 g