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meriva
3 years ago
11

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

Chemistry
2 answers:
Nat2105 [25]3 years ago
8 0

Answer :

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

<u>For 2.72 g KClO_3 :</u>

From the balanced chemical reaction we conclude that,

As, 2\times 122.55 grams of KClO_3 decomposes to give 3\times 32 grams of O_2

So, 2.72 grams of KClO_3 decomposes to give \frac{3\times 32}{2\times 122.55}\times 2.72=1.06 grams of O_2

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

<u>For 0.361 g KClO_3 :</u>

From the balanced chemical reaction we conclude that,

As, 2\times 122.55 grams of KClO_3 decomposes to give 3\times 32 grams of O_2

So, 0.361 grams of KClO_3 decomposes to give \frac{3\times 32}{2\times 122.55}\times 0.361=0.141 grams of O_2

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

<u>For 83.6 kg KClO_3 :</u>

From the balanced chemical reaction we conclude that,

As, 2\times 122.55 grams of KClO_3 decomposes to give 3\times 32 grams of O_2

So, 83.6kg of KClO_3 decomposes to give \frac{3\times 32}{2\times 122.55}\times 83.6=32.7kg of O_2

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

<u>For 22.5 mg KClO_3 :</u>

From the balanced chemical reaction we conclude that,

As, 2\times 122.55 grams of KClO_3 decomposes to give 3\times 32 grams of O_2

So, 22.5mg of KClO_3 decomposes to give \frac{3\times 32}{2\times 122.55}\times 22.5=8.81mg of O_2

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

Diano4ka-milaya [45]3 years ago
4 0
The balanced chemical reaction is given as follows:

<span>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:

</span>2.72 g KClO3 (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 1.06 g O2
<span>
0.361 g KClO3 </span>(1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.14 g O2
<span>
83.6 kg KClO3 (1000g / 1kg) </span>(1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 3275.76 g O2
<span>
22.5 mg KClO3</span> (1 g / 1000 mg) (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.009 g O2
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