What is the limiting reactant when 31.3 g of manganese (II) chloride, 48.3 g of chlorine, and 25.7 g of water react to produce manganese (IV) oxide...

[MandaPaige]

What is the limiting reactant when 31.3 g of manganese (II) chloride, 48.3 g of chlorine, and 25.7 g of water react to produce manganese (IV) oxide and how much hydrochloric acid is produced?

 

When I wrote the balanced equation I got MnCl2 + Cl2 + 2H2O = MnO2 + 4HCl

 

and when I attempted to find how much HCl was produced I came up with 104.2g... which is not an answer choice.

( 25.7*4*36.5/18*2 = 104.2)

 

Can someone check my math or tell me what I am doing wrong? I am also unsure how to go about finding the limiting reactant. This is a review for a final and I'm a little rusty.

 

Any help is appreciated. Thx

[hypervalent_iodine]

That's because you appear to be using mass to determine how much product is formed as opposed to moles. The ratios out the front of the chemical species are mole ratios, not mass ratios. Convert everything to moles work out the limiting reagent first.

Edit: I assume you also used mass to determine limiting reagent? This is also incorrect. Remember that when we talk about things reacting, we're talking about individual molecules colliding to form product. Mass =/= molecules (1 gram of Cl2 will not have the same number of molecules as 1 gram of H2O, for instance) and so it makes no sense to use mass to determine something like a limiting reagent in the way that you have, primarily because the ratios in the reaction are (as mentioned) mole ratios.

[Sensei]

When I wrote the balanced equation I got MnCl2 + Cl2 + 2H2O = MnO2 + 4HCl

 

But reaction is reverse..

MnO₂ + 4 HCl → MnCl₂ + 2 H₂O + Cl₂

 

Preparation page of

https://en.wikipedia.org/wiki/Manganese%28II%29_chloride

Edited December 8, 2015 by Sensei