Question

How many grams of iron are needed to completely consume 51.5 L of chlorine gas according to the following reaction at 25 °C and 1 atm? iron ( s ) + chlorine ( g ) iron(III) chloride ( s )

Answer 1

2 Fe + 3 Cl₂ ----> 2 FeCl₃

PV = nRT

where, P = 1 atm

V = 51.5 L

n = number of moles

R = Gas constant

T = 25 + 273 = 298 K

1 * 51.5 = n * 0.0821 * 298

51.5 = n * 24.5

n = 51.5 / 24.5 = 2.10 mole

Therefore, number of moles of Cl2 = 2.10 mole

From the balanced equation we can say that

3 mole of Cl2 requires 2 mole of Fe so

2.10 mole of Cl2 will require

= 2.10 mole of Cl2 *(2 mole of Fe / 3 mole of Cl2)

= 1.40 mole of Fe

mass of 1 mole of Fe = 55.845 g

so the mass of 1.40 mole of Fe = 78.2 g

Therefore, the mass of Fe required would be 78.2 g