Can someone please help me with these chemistry problem. Thanks.

Chemistry
Tutor: None Selected Time limit: 1 Day

MnO2    +   4  HCl      Cl2    +    MnCl2     +     2 H2O

2.)Given 145.7 grams of manganese (IV) oxide, how much hydrochloric acid (HCl) is needed to use up the MnO2 completely?  

May 19th, 2015

DATA:

Mass of MnO2 = 145.7g

Amount of HCl =?

SOLUTION:

Moles of MnO2 = mass of MnO2/ molecular mass of MnO2

  = 145.7/86.988

  = 1.6749mol of MnO2

1 mol MnO2 = 4 mol HCl

1.6749 mol MnO2 = (4*1.6749) mol HCl

  = 6.6997mol HCl

Mass of HCl = mol of HCl * molecular mass of HCl

  = 6.6997* 36.5

  = 244.5417g HCl


May 19th, 2015

Thank you so much for your help.

May 19th, 2015

my pleasure :)

May 19th, 2015

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