This question is about some of the elements in group 7 of the periodic table - Edexcel - GCSE Chemistry - Question 8 - 2021 - Paper 1

To calculate the percentage by mass of chlorine in POCl₃, we first need to find the molar mass of POCl₃:

• Molar mass of P = 31.0 g/mol
• Molar mass of O = 16.0 g/mol
• Molar mass of Cl = 35.5 g/mol x 3 = 106.5 g/mol

Total molar mass of POCl₃ = 31.0 + 16.0 + 106.5 = 153.5 g/mol.

Now, we can find the percentage by mass of chlorine:

ext{Percentage by mass of Cl} = rac{ ext{mass of Cl}}{ ext{total mass}} imes 100

= rac{106.5}{153.5} imes 100 \\ \approx 69.39\%

First, we will calculate the moles of iron and iron chloride produced in the experiment.

For Reaction A:

• According to the equation, 1 mole of Fe produces 1 mole of FeCl₃.
• Molar mass of Fe = 56.0 g/mol. Thus, moles of Fe = ( \frac{8.40 , ext{g}}{56.0 , ext{g/mol}} = 0.150 \text{ moles} ).
• Therefore, moles of FeCl₃ produced = 0.150 moles.
• Mass of FeCl₃ produced = (0.150 , ext{moles} \times (56.0 + 3 imes 35.5) = 0.150 \times 123.5 = 18.525 , g).

For Reaction B:

• According to the equation, 2 moles of Fe produce 2 moles of FeCl₃.
• Moles of FeCl₃ produced from 0.150 moles of Fe = 0.150 moles.
• Mass of FeCl₃ produced = (0.150 \times 123.5 = 18.525 , g).

Since in the experiment 19.05g of FeCl₃ was formed, the reaction cannot be A.

Now checking reaction B:

• From the equation 2 moles of Fe + 3 moles of Cl₂ produces 2 moles of FeCl₃, which produces:
• Moles of FeCl₃ = (\frac{19.05 , g}{123.5 , g/mol} \approx 0.154 , ext{moles} ).

Thus the excess Fe is appropriate for reaction B, confirming that the reaction which took place was B.