Question

For the following reaction, 22.5 grams of iron are allowed to react with 35.0 grams of chlorine gas . iron(s) chlorine(g) iron(III) chloride(s) What is the maximum mass of iron(III) chloride that can be formed

Answer 1

53.31g of FeCl3

Step-by-step explanation:

The equation of the reaction is given as;

2Fe(s) + 3Cl2(g) --> 2FeCl3(s)

From the stochiometry of the reaction;

2 mol : 3 mol : 2 mol

112g : 213g : 324.4g

Converting the masses given to moles;

Number of moles of Fe = mass / molar mass = 22.5 / 56 = 0.4018

Number of moles of Cl2 = mass / molar mass = 35 / 71 = 0.4930

In this reaction, the limiting reactant is Cl2

213 g of Cl2 produces 324.4g of FeCl3,

35g of Cl2 would produce xg of FeCl3

213 = 324.4

35 = x

x = (35 * 324.4) / 213 = 53.31g