Question

For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.

2 KClO3(s) → 2 KCl(s) + 3 O2(g)
2.72 g KClO3
0.361 g KClO3
83.6 kg KClO3
22.5 mg KClO3

Answer 1

The mass of oxygen form from 2.72 g
`KClO_3` is 1.06 grams.

The mass of oxygen form from 0.361 g
`KClO_3` is 0.141 grams.

The mass of oxygen form from 83.6 kg
`KClO_3` is 32.7 kilograms.

The mass of oxygen form from 22.5 mg
`KClO_3` is 8.81 milligrams.

Explanation :

The given chemical reaction is:

`2KClO_3(s)\rightarrow 2KCl(s)+3O_2(g)`

Molar mass of
`KClO_3` = 122.55 g/mol

Molar mass of
`O_2` = 32 g/mol

For 2.72 g
`KClO_3` :

From the balanced chemical reaction we conclude that,

As,
`2* 122.55` grams of
`KClO_3` decomposes to give
`3* 32` grams of
`O_2`

So,
`2.72` grams of
`KClO_3` decomposes to give
`(3* 32)/(2* 122.55)* 2.72=1.06` grams of
`O_2`

Thus, the mass of oxygen form from 2.72 g
`KClO_3` is 1.06 grams.

For 0.361 g
`KClO_3` :

From the balanced chemical reaction we conclude that,

As,
`2* 122.55` grams of
`KClO_3` decomposes to give
`3* 32` grams of
`O_2`

So,
`0.361` grams of
`KClO_3` decomposes to give
`(3* 32)/(2* 122.55)* 0.361=0.141` grams of
`O_2`

Thus, the mass of oxygen form from 0.361 g
`KClO_3` is 0.141 grams.

For 83.6 kg
`KClO_3` :

From the balanced chemical reaction we conclude that,

As,
`2* 122.55` grams of
`KClO_3` decomposes to give
`3* 32` grams of
`O_2`

So,
`83.6kg` of
`KClO_3` decomposes to give
`(3* 32)/(2* 122.55)* 83.6=32.7kg` of
`O_2`

Thus, the mass of oxygen form from 83.6 kg
`KClO_3` is 32.7 kilograms.

For 22.5 mg
`KClO_3` :

From the balanced chemical reaction we conclude that,

As,
`2* 122.55` grams of
`KClO_3` decomposes to give
`3* 32` grams of
`O_2`

So,
`22.5mg` of
`KClO_3` decomposes to give
`(3* 32)/(2* 122.55)* 22.5=8.81mg` of
`O_2`

Thus, the mass of oxygen form from 22.5 mg
`KClO_3` is 8.81 milligrams.

Answer 2

The balanced chemical reaction is given as follows:

2 KClO3(s) → 2 KCl(s) + 3 O2(g)

The starting amount of the reactant are given above. These values would be used for the calculations. We do as follows:

2.72 g KClO3 (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 1.06 g O2

0.361 g KClO3 (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.14 g O2

83.6 kg KClO3 (1000g / 1kg) (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 3275.76 g O2

22.5 mg KClO3 (1 g / 1000 mg) (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.009 g O2