For the reaction represented by the equation 2kclo3 → 2kcl + 3o2, how many grams of potassium chlorate are required to produce 160 g of oxygen? type the numeric value only.

 The grams  of potassium chlorate  that are required  to produce 160 g of oxygen  is   408.29  grams

calculation

 2 KClO₃→  2 KCl  + 3O₂

Step 1:  find the  moles of  O₂

moles  =  mass÷  molar mass

from periodic table  the  molar mass of O₂  = 16 x2 = 32 g/mol

moles  = 160 g÷ 32  g/mol =  5 moles

Step2 : use the  mole   ratio to determine the  moles of KClO₃

from equation given KClO₃ : O₂  is 2:3

therefore the v moles of KClO₃  =  5 moles x 2/3 = 3.333  moles

Step 3:  find the mass  of KClO₃

mass= moles x molar mass

from periodic table  the molar mass of KClO₃

= 39 + 35.5 + (16 x3) =122.5 g/mol

mass  = 3.333 moles x 122.5 g/mol =408.29 grams

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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