How many moles of O2 are produced from the decomposition of 327g of KCIO3

Explanation:

Potassium chlorate

KClO

3

decomposes to produce potassium chloride

KCl

and oxygen

O

2

. Balancing the equation

KClO

3

→

KCl

+

O

2

(not balanced)

based on the fact that the number of moles of oxygen atoms should be the same on both side of the equation will give

KClO

3

→

KCl

+

3

2

O

2

2

KClO

3

→

2

KCl

+

3

O

2

Hence the ratio of number of moles of particles

n

(

O

2

)

n

(

KClO

3

)

=

3

2

n

(

O

2

)

=

n

(

KClO

3

)

⋅

n

(

O

2

)

n

(

KClO

3

)

=

3

2

⋅

n

(

KClO

3

)

Therefore the decomposition of

6.0

l

mol

of

KClO

3

would yield

6.0

⋅

n

(

O

2

)

n

(

KClO

3

)

=

6.0

⋅

3

2

=

9.0

l

mol

Note that the species

KCl

is highly stable. Both the potassium cation

K

+

and the chloride anion

Cl

−

have attained a noble gas octet configuration of valence shell electrons. Chemical reactions tend to favor the most chemically stable combination. Therefore this decomposition reaction would eventually produce

KCl

rather than other salts that contain oxygen.

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.