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
See the explanationion below
Explanation:
Given:
(a) Mass of
Molecular mass of
No.of moles of
=
Mass of
Molar mass of
No.of moles of
=
Now,
Consider the equation:
Now convert moles to mass,
Mass of
Mass of
As
They can be produced about 188 grams of KCl.
Explanation:
At first, I calculated the mole amount of Cl₂
and KBr:
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.)