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

How many grams of MNCl2 can be produced by 20g of HCl and 10g MnO2?
I need help with the process. I think I go grams to moles of HCl first but am lost

Answer 2

compare (using their stoichiometric coefficients) moles of MnO2 and moles of HCl, whichever has less moles is the limiting reactant.

\(\dfrac{n_{HCl}}{4}\) vs \(\dfrac{n_{MnO_2}}{1}\)

First you have to convert the mass of HCl given to moles use: \(n=\dfrac{m}{M}\)

n=moles, m=mass, M= molar mass

To find moles of HCl (and subsequently, mass), use the moles of the reagent that has more moles (i.e. not the limiting reagent):

e.g. for a general reaction:

\(\color{red}{a}A + \color{blue}{b}B\) \(\rightleftharpoons\) \( \color{green}{c}C\)

where upper case are the species (A,B,C), and lower case (a,b,c) are the coefficients.

\(\dfrac{n_A}{\color{red}{a}}=\dfrac{n_B}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\)

------------------------------------------------------------------------
so if you have moles of B= 2, how many moles of C did you produce?

solve algebraically: \(\dfrac{2}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\rightarrow n_c=\dfrac{2*\color{green}{c}}{\color{blue}{b}}\)
------------------------------------------------------------------------

To convert mass to moles, use the relationship:

\(n=\dfrac{m}{M}\)

where, M=molar mass, m=mass, and n= moles.