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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?,

Respuesta :

W0lf93
Given 
Mole of MnO2 is 0.86
 mass of HCl is 48.2 g 
Molarity of HCl is 36.5 g/mole,
 Molarity of Cl2 is 71 g/mole  
mole of HCl = mass of HCl/Molarity of HCl = 48.2/36.5 = 1.32 mole 
From the above chemical equation: 4 moles of HCl reacts with 1 mole of
MnO2 . Thus,
0.86 mole of MnO2 require 4·0.86 = 3.44 moles of HCl to react.
but only 1.32 mole are present (as calculated earlier).
This implies HCl is in lesser amount compared with MnO2 and
Hence, HCl will be used up first. 

Now, From the above given chemical equation: 
1 mole of Cl2 is produced when 4 moles of HCl react. So, 
4 mole * 36.5 g/mole HCl –---> 1 mole * ·71g/mole Cl2 
48.2 g HCl –----> mass of Cl2  
mass of Cl2 = 71 * 48.2 / (4 *36.5) = 23.4 g

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