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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

Verified Answer :

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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

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

For 2.72 g \(KClO_3\) :

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.

For 0.361 g \(KClO_3\) :

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.

For 83.6 kg \(KClO_3\) :

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.

For 22.5 mg \(KClO_3\) :

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.

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