If 34.2g of KMnO4 were mixed with 0.990 moles of HCL, how many moles of Chlorine gas could be produced?

Question

Can someone tell me how to solve this problem please?

2 KMnO4+16 HCL ---> 2 MnCl2+ 2 KCL+ 5 Cl2 + 8 H20

Answer 1

Mole KMnO4 = 34.2 g / 158 = 0.216

the ratio is 2 : 16

2 : 16 = x : 0.990

x = 0.124 moles KMnO4 needed for the reaction

we have 0.216 moles so KMnO4 is in excess and HCl is the limiting reactant

the ratio between HCl and Cl2 is 16 : 5

16 : 5 = 0.990 : x

x = 0.309 moles Cl2

0.309 mol x 70.9 g/mol = 21.9 g

Answer 2

Calculate the mol. wt of KMnO4 first.