How many grams of POCl3 are produced when 225.0 grams of P4O10 and 675.0 grams of PCl5 react? This is

3 answers
Question:

How many grams of POCl3 are produced when 225.0 grams of P4O10 and 675.0 grams of PCl5 react? This is the balance equation P4O10 + 6PCl5 → 10POCl3

Answers

900g of POCl₃

Explanation:

Hello,

To solve this question, we'll require the equation of reaction.

P₄O₁₀ + 6PCl₅ → 10POCl₃

Molar mass of P₄O₁₀ = 283.886 g/mol

Molar mass of PCl₅ = 208.24 g/mol

Molar mass of POCl₃ = 153.33 g/mol

But Number of moles = mass / molar mass

Mass = molar mass × number of moles

Mass of POCl₃ = 153.33 × 10 = 1533.3g

Mass of PCl₅ = 208.24 × 6 = 1249.44g

Mass of P₄O₁₀ = 283.886 × 1 = 283.886g

From the equation of reaction,

283.886g of P₄O₁₀ + 1249.44g of PCl₅ produces 1533.33g of POCl₃

I.e 1533.33g of reactants produces 1533.33g of product (law of conservation of mass)

Therefore, (225g of P₄O₁₀ + 675g of PCl₅) = 900g will give x g of POCl₃.

1533.33g of reactants = 1533.33g of products

900g of reactants = x g of products

x = (900 × 1533.33) / 1533.33

x = 900g of POCl₃

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