Two moles of calcium phosphate on reduction with carbon in the presence of silica resulted in the formation of a phosphorus compound X in 90% yield. The weight of X is (Atomic weight, Ca 40, P 31, Si 28, O 16, C 12, H 1)

Explanation: 

\(2 \text{Ca}_3(\text{PO}_4)_2 + 10 \text{C} + 6 \text{SiO}_2 \rightarrow \text{P}_4 + 6 \text{CaSiO}_3 + 10 \text{CO}\)

• Step 1: Molar masses of the relevant compounds:

  • Molar mass of (Ca₃(PO₄)₂):

    • \(3 \times \text{Ca} + 2 \times \text{P} + 8 \times \text{O} = 3 \times 40 + 2 \times 31 + 8 \times 16 = 310 \, \text{g/mol} \)

  • Molar mass of (P₄):

    • \( 4 \times \text{P} = 4 \times 31 = 124 \, \text{g/mol} \)

• Step 2: Stoichiometric relationship between calcium phosphate and phosphorus ((P₄)):

  • From the balanced equation, 2 moles of (Ca₃(PO₄)₂) produce 1 mole of (P₄).

  • Therefore, if 2 moles of calcium phosphate are used, we get 1 mole of (P₄), which weighs 124 g.

• Step 3: Account for the 90% yield of the reaction:

  \(\text{Actual weight of } \text{P}_4 = \text{Theoretical weight of } \text{P}_4 \times \text{Yield} = 124 \, \text{g} \times 0.9 = 111.6 \, \text{g}\)

  • Since the yield of the reaction is 90%, we need to calculate the actual amount of (P₄) produced:

Conclusion:

The actual weight of phosphorus compound X produced is 111.6 g.