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how many milligrams of iron(iii) chloride result when 13.1 mg of iron is reacted with an excess of chlorine gas?

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Answer: 38.1 mg

Step-by-step explanation:

Step 1: Write the equation for the reaction.

Fe (s) + Cl₂ (g) ⇒ FeCl₃ (s)

Step 2: Balance the equation.

2 Fe (s) + 3 Cl₂ (g) ⇒ 2 FeCl₃ (s)

Step 3: Convert 13.2 mg iron into moles iron.

13.1 mg x (1 g/1000 mg) x (1 mol/55.8 g) = .000235 mol Fe

Step 4: Use the balanced equation from step 2, the moles of Fe from step 3, and the molar mass of FeCl₃ to find the moles FeCl₃.

1 mole FeCl₃ = 162.2 g/mol

.000235 mol Fe x (2 mol FeCl₃/2 mol Fe) x (162.2 g FeCl₃/1 mol FeCl₃) = 0.0381 g FeCl₃

Step 5: Convert grams FeCl₃ to milligrams FeCl₃.

1 g = 1000 mg

0.0381 g FeCl₃ x 1000 mg/1 g = 38.1 mg

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