Iron metal reacts with chlorine gas according to the following equation: 2Fe(s) + 3Cl₂(g) $\rightarrow$2FeCl₃(s) If 35.0 g each of iron and chlorine are combined, how much FeCl₃ should form?

Question

Quizplus · Accepted Answer

To solve this problem, we need to use stoichiometry. First, we need to determine which reactant is limiting. We can do this by calculating the amount of product that would be formed if all of the iron and all of the chlorine reacted completely.

Molar mass of Fe = 55.85 g/mol
Molar mass of Cl2 = 70.90 g/mol

Number of moles of Fe = 35.0 g / 55.85 g/mol = 0.626 mol
Number of moles of Cl2 = 35.0 g / 70.90 g/mol = 0.494 mol

According to the balanced chemical equation, 2 moles of Fe react with 3 moles of Cl2 to produce 2 moles of FeCl3. Therefore, if 0.626 moles of Fe react completely, they would require 0.939 moles of Cl2 (since the ratio is 2:3). This amount of Cl2 is greater than the 0.494 moles we have available, so Cl2 is the limiting reactant.

Now we can use the amount of limiting reactant to determine the amount of product formed. Using the same mole ratios as above, we can calculate:

Number of moles of FeCl3 = 0.494 mol Cl2 x (2 mol FeCl3 / 3 mol Cl2) = 0.329 mol FeCl3

Finally, to convert moles of FeCl3 to grams, we use the molar mass of FeCl3:

Molar mass of FeCl3 = 162.20 g/mol
Mass of FeCl3 formed = 0.329 mol x 162.20 g/mol = 53.4 g

Therefore, the answer is C) 53.4 g.

Iron Metal Reacts with Chlorine Gas According to the Following →\rightarrow→

Iron Metal Reacts with Chlorine Gas According to the Following $\rightarrow$