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
mols K3PO4 = grams/molar mass = ?
mols H3PO4 used in the reaction is 1 mol H3PO4 used for every 1 mol K3PO4 produced (look at the coefficients).
grams H3PO4 to start = trams used + grams remaining
To determine the starting mass of each reactant, we need to use stoichiometry - the molar ratio between the reactants and products.
First, we need to write and balance the chemical equation:
3KOH + H3PO4 ----> K3PO4 + 3H2O
Next, we can calculate the molar mass of each compound using the periodic table:
- Molar mass of KOH = atomic mass of K (39.1) + atomic mass of O (16.0) + atomic mass of H (1.0) = 56.1 g/mol
- Molar mass of H3PO4 = atomic mass of H (1.0) + atomic mass of P (31.0) + atomic mass of O (16.0) x 4 = 98.0 g/mol
- Molar mass of K3PO4 = atomic mass of K (39.1) x 3 + atomic mass of P (31.0) + atomic mass of O (16.0) x 4 = 212.3 g/mol
Now, we can calculate the number of moles of K3PO4 produced using its mass:
- Moles of K3PO4 = mass of K3PO4 / molar mass of K3PO4
= 46.3 g / 212.3 g/mol
≈ 0.218 mol
Since the stoichiometric ratio between KOH and K3PO4 is 3:1, we can determine the number of moles of KOH required:
- Moles of KOH = moles of K3PO4 x (3 moles of KOH / 1 mole of K3PO4)
≈ 0.218 mol x (3/1)
≈ 0.654 mol
Finally, we can calculate the mass of KOH using its molar mass:
- Mass of KOH = moles of KOH x molar mass of KOH
= 0.654 mol x 56.1 g/mol
≈ 36.7 g
Therefore, the starting mass of KOH is approximately 36.7 grams.
Similarly, we can determine the starting mass of H3PO4 by subtracting the mass of K3PO4 produced from the initial mass of H3PO4:
- Mass of H3PO4 = initial mass of H3PO4 - mass of K3PO4 produced
= 92.8 g - 46.3 g
≈ 46.5 g
Therefore, the starting mass of H3PO4 is approximately 46.5 grams.