Question

oxygen is produced by catalytic decomposition of KClO3 . 2KCl⇒⇒2KCl + 3O2.

calculate the amount of KClO3 required to produce 2.4 moles of oxygen

Answer 1

The equation is as follows:
             
   
                           MnO2
         2KCLO3 ------------------> 2KCl  + 3O2

The mole ratio for O2:KCLO3 is 3:2

Therefore 2.4 moles of oxygen requires:

if 3 = 2.4 moles
Then 2= 2.4 x 2/3
  
           =  0.8 x 2 = 1.6 moles

Moles = mass/molar mass
mass= moles x molar mass

moles= 1.6
molar mass of KClO3= 122.5

Therefore mass = 1.6 x 122.5
                          = 196 g

Answer 2

Answer: The amount of KClO3 required 196 g.  

Given:

196 g of $KCl{ O }_{ 3 }$ is requires to produce 2.4 moles of oxygen.

Solution:

The given chemical reaction is as follows

$2KCl{ O }_{ 3 }\quad \xrightarrow { Mn{ O }_{ 2 } } \quad 2KCl\quad +\quad 3{ O }_{ 2 }$

From the reaction,

3 moles of oxygen requires $KCl{ O }_{ 3 }$ = 2 moles

$So,\quad 1\quad mole\quad of\quad oxygen\quad will\quad require\quad KCl{ O }_{ 3 }\quad =\quad \frac { 2 }{ 3 } \quad mole$

$So,\quad 2.4\quad moles\quad of\quad oxygen\quad will\quad require\quad KCl{ O }_{ 3 }\quad =\quad \frac { 2 }{ 3 }\quad \times \quad 2.4\quad =\quad 1.6\quad moles$

$Moles\quad =\quad \frac { Mass\quad  }{ Molar\quad mass }$

$Mass\quad =\quad moles\quad \times \quad molar\quad mass$

$Moles\quad =\quad 1.6$

$Molar\quad mass\quad of\quad KCl{ O }_{ 3 }\quad =\quad 122.5$

Therefore, mass of $KCl{ O }_{ 3 }$ required to produce 2.4 moles of oxygen $= 1.6\quad \times \quad 122.5\quad =\quad 196\quad g$