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.

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.