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An impurity sometimes found in Ca₃(PO₄)₂ is Fe₂O₃, which is removed during the production of phosphorus as ferrophosphorus (Fe₂P).(b) If 50. metric tons of crude Ca₃(PO₄)₂ contains 2.0% Fe₂O₃ by mass and the overall yield of phosphorus is 90.%, how many metric tons of P₄ can be isolated?

Respuesta :

shaikhmohd0119

Actual amount of the phosphorous produced is 8.8 metric ton

  • Mass of crude Ca₃(PO₄)₂ mineral is 50 tons

percentage of Fe2O3 = 2% , so percentage of pure Ca₃(PO₄)₂= 98%

actual amount of  Ca₃(PO₄)₂:

98% (50t) = 98/100 (50t)

                = 49t × 10⁶ g/1 t

                =4.9 × 10⁷ g

  • Molar mass of Ca₃(PO₄)₂ = 311.2g/mol (40×3 + (31+16×4)×2)
  • no of moles of  Ca₃(PO₄)₂= 4.9×10⁷ g × 1mol Ca₃(PO₄)₂/311.2 g

                                          = 1.57 × 10⁵ mol

  • The equation representing the production of P₄ from  Ca₃(PO₄)₂ is :

2Ca₃(PO₄)₂ (s) + 6SiO₂(s) + 10 C (s) → P₄(s) + 6 CaSiO₃ (s) + 10CO (g)

According to equation: 2 mol of Ca₃(PO₄)₂  produce 1 mol P₄

Molar mass of P₄ is 123.9 g/mol ( 30.97×4)

  • 2 mol   Ca₃(PO₄)₂ = 1 mol P₄

         1 mol      P₄          = 123.9 g

          1 t  = 10⁶ g

  • Theoretical yield of phosphorus:

 1.57 × 10⁵ mol  Ca₃(PO₄)₂ × (1 mol P₄ / 2 mol Ca₃(PO₄)₂)× (123.9g/1 mol P₄)× 1 t/ 10⁶ g = 9.73 t

Percent yield of reaction = 90%

  • Percent yield = (actual yield / theoretical yield ) × 100 %

        Actual yield = (percent yield × theoretical yield ) / 100 %

                             = 90% ×9.73t /100%

                             = 8.76 t ≈ 8.8 t

Therefore, actual amount of the phosphorous produced is 8.8 metric ton

Thus we can conclude that 8.8 metric tons of P₄ can be isolated

learn more to calculate moles at https://brainly.com/question/13314627

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