4. Phosphorous trichloride, PCl3, reacts with water to form phosphorous acid, H3PO4, and hydrocholoric acid:
PCl3+ 3H2O> H3PO3+3HCL
a. what is the limiting reactant when 10.0g of PCl3 is mixed with 12.4 g of H2O?
b.what mass of HCl is formed?
C. how much excess reactant remains after the reaction stops?
same thing.
To determine the limiting reactant, you need to compare the amount in moles of each reactant and calculate the number of moles of the products that can be formed from each reactant. The reactant that produces the smaller number of moles of product is the limiting reactant.
a. Start by converting the mass of each reactant to moles. You can do this by using the molar mass of each substance.
Molar mass of PCl3:
P (1 atom) = 1.0 g/mol
Cl (3 atoms) = 3 × 35.45 g/mol = 106.35 g/mol
Total molar mass = 107.35 g/mol
Molar mass of H2O:
H (2 atoms) = 2 × 1.01 g/mol = 2.02 g/mol
O (1 atom) = 1 × 16.00 g/mol = 16.00 g/mol
Total molar mass = 18.02 g/mol
Now, calculate the number of moles for each reactant:
moles of PCl3 = mass of PCl3 / molar mass of PCl3
= 10.0 g / 107.35 g/mol
≈ 0.093 moles
moles of H2O = mass of H2O / molar mass of H2O
= 12.4 g / 18.02 g/mol
≈ 0.687 moles
Next, balance the chemical equation:
PCl3 + 3H2O → H3PO3 + 3HCl
According to the balanced equation, the ratio of moles of PCl3 to moles of H2O is 1:3. Therefore, for every mole of PCl3, 3 moles of H2O are required.
Now, calculate the number of moles of product that can be formed from each reactant:
moles of H3PO3 (from PCl3) = moles of PCl3 × (1 mole H3PO3 / 1 mole PCl3)
≈ 0.093 moles
moles of H3PO3 (from H2O) = moles of H2O × (1 mole H3PO3 / 3 moles H2O)
≈ 0.229 moles
Since the moles of H3PO3 (from PCl3) is lower than the moles of H3PO3 (from H2O), PCl3 is the limiting reactant. Therefore, the limiting reactant is PCl3.
b. To determine the mass of HCl formed, you need to calculate the number of moles of HCl produced using the balanced chemical equation and then convert it to grams using the molar mass of HCl.
From the balanced equation, you can see that for every mole of PCl3, 3 moles of HCl are produced.
moles of HCl = moles of PCl3 × (3 moles HCl / 1 mole PCl3)
= 0.093 moles × 3
= 0.279 moles
mass of HCl = moles of HCl × molar mass of HCl
= 0.279 moles × 36.46 g/mol
≈ 10.17 g
Therefore, approximately 10.17 grams of HCl are formed.
c. To determine how much excess reactant remains after the reaction stops, subtract the moles of the limiting reactant used from the initial moles of the excess reactant.
moles of H2O remaining = moles of H2O initially - (moles of PCl3 × (3 moles H2O / 1 mole PCl3))
= 0.687 moles - (0.093 moles × 3)
= 0.498 moles
mass of H2O remaining = moles of H2O remaining × molar mass of H2O
= 0.498 moles × 18.02 g/mol
≈ 8.97 g
Therefore, approximately 8.97 grams of H2O remains as excess reactant after the reaction stops.