if you are provided with 34.5 g iron (iii) chloride and 25.3 g sodium carbonate what is the limiting ractant

Sodium carbonate, Na₂CO₃ is the limiting

Explanation:

We'll begin by writing the balanced equation for the reaction between iron (iii) chloride, FeCl₃ and sodium carbonate, Na₂CO₃.

This is illustrated below:

2FeCl₃ + 3Na₂CO₃ → Fe₂(CO₃)₃ + 6NaCl

Next, we shall determine the masses of FeCl₃ and Na₂CO₃ that reacted from the balanced equation. This can be obtained as follow:

Molar mass of FeCl₃ = 56 + (3×35.5)

= 56 + 106.5

= 162.5 g/mol

Mass of FeCl₃ from the balanced equation = 2 × 162.5 = 325 g

Molar mass of Na₂CO₃ = (2×23) + 12 + (3×16)

= 46 + 12 + 48

= 106 g/mol

Mass of Na₂CO₃ from the balanced equation = 3 × 106 = 318 g

SUMMARY:

From the balanced equation above,

325 g of FeCl₃ reacted with 318 g of Na₂CO₃.

Finally, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

325 g of FeCl₃ reacted with 318 g of Na₂CO₃.

Therefore, 34.5 g of FeCl₃ will react with = (34.5 × 318)/325 = 33.76 g of Na₂CO₃.

From the calculations made above, we can see clearly that it will take a higher amount (i.e 33.76 g) of Na₂CO₃ than what was given (i.e 25.3 g) to react completely with 34.5 g of FeCl₃. Therefore, Na₂CO₃ is the limiting reactant and FeCl₃ is the excess reactant.

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