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Phosphorus can exist as P₄ molecules - OCR Gateway - GCSE Chemistry - Question 21 - 2022 - Paper 3

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Phosphorus can exist as P₄ molecules. Phosphorus trichloride, PCl₃, is made in the reaction of phosphorus, P₄, and chlorine as shown in the equation. P₄ + 6Cl₂ → 4... show full transcript

Worked Solution & Example Answer:Phosphorus can exist as P₄ molecules - OCR Gateway - GCSE Chemistry - Question 21 - 2022 - Paper 3

Step 1

Calculate the mass of phosphorus.

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Answer

To calculate the mass of 2.0 mol of phosphorus (P₄), we first need to find the molar mass of P₄:

  1. The atomic mass of phosphorus (P) is approximately 31 g/mol.

  2. The molecular formula for tetraphosphorus is P₄, thus:

    extMolarmassofP4=4imes31extg/mol=124extg/mol ext{Molar mass of P₄} = 4 imes 31 ext{ g/mol} = 124 ext{ g/mol}

  3. Next, we calculate the total mass for 2.0 mol:

    extMass=extmolesimesextmolarmass=2.0extmolimes124extg/mol=248extg ext{Mass} = ext{moles} imes ext{molar mass} = 2.0 ext{ mol} imes 124 ext{ g/mol} = 248 ext{ g}

Therefore, the mass of phosphorus is 248 g.

Step 2

Calculate the maximum mass of phosphorus trichloride.

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Answer

To determine the maximum mass of phosphorus trichloride (PCl₃) made from 2.0 mol of phosphorus (P₄), we need to analyze the balanced chemical equation:

  1. The reaction indicates that 1 mole of P₄ produces 4 moles of PCl₃.

  2. Therefore, from 2.0 mol of P₄, the moles of PCl₃ produced will be:

    extMolesofPCl3=2.0extmolP4imes4=8.0extmolPCl3 ext{Moles of PCl₃} = 2.0 ext{ mol P₄} imes 4 = 8.0 ext{ mol PCl₃}

  3. We now calculate the molar mass of PCl₃:

    • Molar mass of chlorine (Cl) is approximately 35.5 g/mol, thus:

    extMolarmassofPCl3=31extg/mol(P)+3imes35.5extg/mol(Cl)=31+106.5=137.5extg/mol ext{Molar mass of PCl₃} = 31 ext{ g/mol (P)} + 3 imes 35.5 ext{ g/mol (Cl)} = 31 + 106.5 = 137.5 ext{ g/mol}

  4. Finally, we find the maximum mass of PCl₃ produced:

    extMaximummass=extmolesimesextmolarmass=8.0extmolimes137.5extg/mol=1100extg ext{Maximum mass} = ext{moles} imes ext{molar mass} = 8.0 ext{ mol} imes 137.5 ext{ g/mol} = 1100 ext{ g}

Thus, the maximum mass of phosphorus trichloride produced is 1100 g.

Step 3

Which is the limiting reactant? Explain your answer.

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Answer

To identify the limiting reactant, we assess the quantities of both reactants:

  1. Starting moles of P₄:

    • We have 2.0 mol of phosphorus (P₄).
  2. Chlorine needed for the reaction:

    • According to the balanced equation, 6 moles of Cl₂ are required for every 1 mole of P₄. Thus:

    extMolesofCl2needed=2.0extmolP4imes6=12.0extmolCl2 ext{Moles of Cl₂ needed} = 2.0 ext{ mol P₄} imes 6 = 12.0 ext{ mol Cl₂}

  3. Available chlorine:

    • The given amount of chlorine is 866.2 g. The moles of Cl₂ can be calculated by:

ightarrow rac{866.2 ext{ g}}{35.5 ext{ g/mol}} ext{ ≈ 24.4 mol}$$

  1. Since only 12.0 moles of Cl₂ are needed for 2.0 moles of P₄, P₄ is the limiting reactant.

    Thus, the limiting reactant is phosphorus (P₄), because it produces a smaller amount of product compared to chlorine.

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