Phosphorus(III) chloride, PCl3 is used in the manufacture of pesticides. It is made by the direct combination of phosphorus & chloride. Calculate the mass of PCl3 produced when
125 grams of P4 was allowed to react with 323 grams of chlorine gas?
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To calculate the mass of PCl3 produced, we need to determine the limiting reactant first. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
Let's start by calculating the number of moles of each reactant.
1. Moles of P4:
Given mass of P4 = 125 grams
Molar mass of P4 = 4 * 31.0 g/mol = 124.0 g/mol
Number of moles of P4 = Mass / Molar mass
Number of moles of P4 = 125 g / 124.0 g/mol = 1.008 moles
2. Moles of Cl2:
Given mass of Cl2 = 323 grams
Molar mass of Cl2 = 2 * 35.5 g/mol = 71.0 g/mol
Number of moles of Cl2 = Mass / Molar mass
Number of moles of Cl2 = 323 g / 71.0 g/mol = 4.56 moles
Now we need to determine the stoichiometry of the reaction. The balanced equation for the reaction is:
P4 + 6Cl2 -> 4PCl3
According to the balanced equation, each mole of P4 reacts with 6 moles of Cl2 to produce 4 moles of PCl3.
From the stoichiometry, we can see that the mole ratio of P4 to Cl2 is 1:6. This means that for every mole of P4, we need 6 moles of Cl2.
Since we have 1.008 moles of P4 and 4.56 moles of Cl2, we can calculate how many moles of PCl3 can be produced using the limiting reactant (P4).
Number of moles of PCl3 produced = (Number of moles of P4) * (Mole ratio of PCl3 to P4)
Number of moles of PCl3 produced = 1.008 mol * (4 mol PCl3 / 1 mol P4) = 4.032 mol
Now, we can calculate the mass of PCl3 produced using the number of moles and the molar mass of PCl3.
Molar mass of PCl3 = 31.0 g/mol + 3 * 35.5 g/mol = 137.5 g/mol
Mass of PCl3 produced = Number of moles of PCl3 * Molar mass of PCl3
Mass of PCl3 produced = 4.032 mol * 137.5 g/mol = 555.36 grams
Therefore, the mass of PCl3 produced when 125 grams of P4 reacts with 323 grams of Cl2 is 555.36 grams.
To calculate the mass of PCl3 produced, we first need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed, thus limiting the amount of product formed.
To find the limiting reactant, we compare the moles of each reactant to the stoichiometric ratio in the balanced chemical equation:
1) Calculate the number of moles of P4:
Number of moles = mass / molar mass
Molar mass of P4 = 4 * atomic mass of phosphorus = 4 * 31.0 g/mol = 124.0 g/mol
Number of moles of P4 = 125 g / 124.0 g/mol = 1.008 moles
2) Calculate the number of moles of Cl2:
Number of moles = mass / molar mass
Molar mass of Cl2 = 2 * atomic mass of chlorine = 2 * 35.5 g/mol = 71.0 g/mol
Number of moles of Cl2 = 323 g / 71.0 g/mol = 4.563 moles
3) Calculate the ratio of moles of P4 to Cl2:
P4 : Cl2 = 1.008 moles : 4.563 moles ≈ 1 : 4.521 (rounded to three decimal places)
Since the ratio is close to 1:4, it suggests that the stoichiometric ratio in the balanced equation is 1:4, meaning 1 mole of P4 reacts with 4 moles of Cl2 to produce 4 moles of PCl3.
4) Determine the moles and mass of PCl3 produced:
From the above calculation, we know that for every mole of P4, 4 moles of PCl3 are produced. Since we have already determined that P4 is the limiting reactant, we can directly calculate the moles and then the mass of PCl3 produced.
Number of moles of PCl3 = 4 * 1.008 moles = 4.032 moles
Molar mass of PCl3 = atomic mass of phosphorous + 3 * atomic mass of chlorine = 31.0 + 3 * 35.5 = 137.5 g/mol
Mass of PCl3 produced = number of moles * molar mass = 4.032 moles * 137.5 g/mol ≈ 554.64 grams (rounded to two decimal places)
Hence, approximately 554.64 grams of PCl3 will be produced when 125 grams of P4 reacts with 323 grams of chlorine gas.
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