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What mass of PCl₅ will be produced from the given masses of both reactants?

28.0 g of P₄ and 59.0 g of Cl₂

1 Answer

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Final answer:

The mass of PCl₅ produced is calculated using stoichiometry from the balanced chemical equation, which indicates that 28.0 grams of P₄ reacts with 59.0 grams of Cl₂ to produce 188.25 grams of PCl₅, with P₄ as the limiting reagent.

Step-by-step explanation:

To calculate the mass of PCl₅ produced from 28.0 g of P₄ and 59.0 g of Cl₂, you need the balanced chemical equation for the reaction. However, the provided information does not include this equation, so let's assume it's as follows:

P₄(s) + 10 Cl₂(g) → 4 PCl₅(s)

First, we calculate the moles of P₄ and Cl₂ based on the given masses and their molar masses:

  • Molar mass of P₄: 4 x 30.974 g/mol = 123.896 g/mol
  • Molar mass of Cl₂: 2 x 35.45 g/mol = 70.90 g/mol
  • Moles of P₄: 28.0 g / 123.896 g/mol = 0.226 moles
  • Moles of Cl₂: 59.0 g / 70.90 g/mol = 0.832 moles

From the stoichiometry of the balanced equation, we can see that 1 mole of P₄ reacts with 10 moles of Cl₂ to produce 4 moles of PCl₅. Thus, P₄ is the limiting reagent here since it will run out first:

  • Moles of PCl₅ produced: 0.226 moles P₄ x (4 moles PCl₅ / 1 mole P₄) = 0.904 moles PCl₅
  • Mass of PCl₅ produced: 0.904 moles x 208.24 g/mol (molar mass of PCl₅) = 188.25 g

The mass of PCl₅ produced from the reaction is 188.25 g, assuming that P₄ is the limiting reagent.

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