Question

If 3.18 x 10^23 atoms of iron react with 67.2 L of chlorine gas at STP, what is the maximum

Answer 1

84.34 grams of grams of iron (III) chloride that can be produced is maximum because Fe is the limiting reagent in this reaction and chlorine gas is excess reagent.

Explanation:

Balanced chemical equation:

2 Fe + 3 Cl2 → 2 FeCl3​

DATA GIVEN:

iron =  atoms

mass of chlorine gas = 67.2 liters

mass of FeCl3 = ?

number of moles of iron will be calculated as

number of moles = $\frac{total number of atoms}{Avagaro's number}$

number of moles = $\frac{3.18 x 10^23}{6.022x 10^23}$

number of moles = 0.52 moles of iron

moles of chlorine gas

number of moles = $\frac{mass}{molar mass of 1 mole}$

Putting the values in the equation:

n = $\frac{67200}{70.96}$               (atomic mass of chlorine gas = 70.96 grams/mole)

   = 947.01 moles

Fe is the limiting reagent so

2 moles of Fe gives 2 moles of FeCl3

0.52 moles of Fe will give

$\frac{2}{2}$ = $\frac{x}{0.52}$

0.52 moles of FeCl3 is formed.

to convert it into grams:

mass = n X atomic mass

         = 0.52 x 162.2                   (atomic mass of FeCl3 is 162.2grams/mole)  

          = 84.34 grams