Question

MnO2+ 4HCL--->MnCl2+Cl2+2H2O:
if 0.86 mole of MnO2 and 48.2g of HCL react, which reagent will be used up first? How many grams of Cl2 will be produced?
The answers are HCL, and 23.4g.
How would I get these answers?

Answer 1

first you need to find the limiting reagent
given that moles of MnO2 = 0.86 moles

we know that moles = weight / molecular weight

amount of MnO2 = moles * molecular weight
amount of MnO2 = 0.86 * 86.9 = 74.7 g

this is more than the amount of HCl present .(amount of HCl = 48.2 g)

Answer 2

from the stoichiometric eqn we can say that
4 moles of HCl reacts to give 1 mole of Cl2
from the given data
amount of HCl = 48.2 g
number of moles of HCl = 48.2/36.5

now 48.2/36.5 moles of HCl gives ------- x moles of Cl2

multiply the value of x with molecular weight of Cl2 (mol.wt = 71)
you will the above answer...