Question

How many grams of iron are needed to completely consume 37.4 L of chlorine gas according to the following reaction at 25 °C and 1 atm? iron (s) + chlorine (g)- iron(III) chloride (s) grams iron

Answer 1

mass of iron needed = 56.9 g

Explanation

The balanced reaction equation is : 2 Fe + 3 Cl₂ $\rightarrow$ 2 FeCl₃

volume of Cl₂ = 37.4 L

moles Cl₂ = [(pressure Cl₂) * (volume of Cl₂)] / [(R) * (temperature Cl₂)]

where R = 0.0821 L-atm/mol-K

moles Cl₂ = [(1 atm) * (37.4 L)] / [(0.0821 L-atm/mol-K) * (298 K)]

moles Cl₂ = 1.53 mol

moles Fe = (moles Cl₂) * (2 moles Fe / 3 moles Cl₂)

moles Fe = (1.53 mol) * (2/3)

moles Fe = (1.53 mol) * (0.666)

moles Fe = 1.02 mol

mass Fe = (moles Fe) * (molar mass Fe)

mass Fe = (1.02 mol) * (55.845 g/mol)

mass Fe = 56.9 g