For the reaction represented by the equation Cl_2 + 2KBr -> 2KCl + Br_2, how many grams of potassium chloride can be produced from 300. g each of chlorine and potassium bromide?

2 Answers
Apr 13, 2017

See the explanationion below:

Explanation:

Cl_2 + 2KBr -> 2KCl + Br_2
Given:
300g of Cl_2
300g of KBr

(a) Mass of Cl_2 = 300g
Molecular mass of Cl_2 = 2(35.45) = 70.906 g/mol
No.of moles of Cl_2 = mass/molar mass = 300/70.906
=4.23 moles
Mass of KBr = 300g
Molar mass of KBr= 39+79.9 = 118.9g/mol
No.of moles of KBr = 300/118.9
=2.52moles
Now,
Consider the equation:
Cl_2 +2KBr-> 2KCl + Br_2
1 mole of Cl_2 = 2 moles of KCl
4.23 moles of Cl_2 = 2x4.23= 8.46 moles

2 moles of KBr = 2 moles of KCl
2.52 moles of KBr = 2.52 moles

Now convert moles to mass,
Mass of KCl produced by Cl_2 = 8.46×70.906=599.8g
Mass of KCl produced by potassium bromide = 2.52×74.5=187.74g

As KBr is producing lesser moles of product so, KBr is limiting reactant and produced potassium chloride= 187.74g

Apr 15, 2017

They can be produced about 188 grams of KCl.

Explanation:

At first, I calculated the mole amount of Cl₂

"300 g"/"35.45 g/mol" = 8.463\ mol"

and KBr:

"300 g"/"119.01 g/mol" = 2.521\ mol.

Then, by observing that half of dichlorine moles (4.232 mol) are greater than the available moles of KBr, I deduced potassium bromide is the limiting reactant.

Consequently, provided the stoichiometric coefficient of the required product, KCl, is the same of KBr, 2.521 moles of KBr will be converted in the same mole amount of KCl.

Eventually, I calculated the mass of 2.521 moles of KCl:

2.521 mol · 74.55 g/mol = 187.9 g.

(by student Alessia, checked by prof. a.t.)