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3. If 252 grams of iron are reacted with 321 grams of chlorine gas, how many grams of FeCl3 are produced by this reaction?2Fe(s) + 3Cl2(g) → 2FeCl3(s)

Question

3. If 252 grams of iron are reacted with 321 grams of chlorine gas, how many grams of FeCl3 are produced by this reaction?

2Fe(s)+3Cl2(g)2FeCl3(s) 2Fe(s) + 3Cl_2(g) \rightarrow 2FeCl_3(s)

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Solution

First, we need to determine the molar mass of each substance involved in the reaction.

The molar mass of Fe (iron) is approximately 55.85 g/mol. The molar mass of Cl2 (chlorine gas) is approximately 70.90 g/mol (35.45 g/mol for each Cl atom). The molar mass of FeCl3 (iron(III) chloride) is approximately 162.20 g/mol (55.85 g/mol for Fe and 35.45 g/mol for each of the three Cl atoms).

Next, we need to determine the number of moles of each substance.

For iron, we have 252 g / 55.85 g/mol = approximately 4.51 moles. For chlorine gas, we have 321 g / 70.90 g/mol = approximately 4.52 moles.

According to the balanced chemical equation, 2 moles of Fe react with 3 moles of Cl2 to produce 2 moles of FeCl3.

Therefore, the reaction is limited by the amount of iron, as we have slightly less iron (4.51 moles) than chlorine gas (4.52 moles).

So, we can expect to produce 2 moles of FeCl3 for every 2 moles of Fe, or in other words, 1 mole of FeCl3 for every 1 mole of Fe.

Therefore, we can expect to produce approximately 4.51 moles of FeCl3.

Finally, to find the mass of FeCl3 produced, we multiply the number of moles by the molar mass:

4.51 moles * 162.20 g/mol = approximately 731.62 g of FeCl3.

So, we can expect to produce approximately 731.62 g of FeCl3 from 252 g of Fe and 321 g of Cl2.

This problem has been solved

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