Question

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

number of grams of iron (III) chloride that can be produced?

2 Fe + 3 Cl2 → 2 FeCl3​

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.

Step-by-step 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 =
`(total number of atoms)/(Avagaro's number)`

number of moles =
`(3.18 x 10^23)/(6.022x 10^23)`

number of moles = 0.52 moles of iron

moles of chlorine gas

number of moles =
`(mass)/(molar mass of 1 mole)`

Putting the values in the equation:

n =
`(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

`(2)/(2)` =
`(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