How many grams of potassium chlorate should be decomposed to obtain 9.6g of oxygen? (Atomic mass of K = 39 u, Cl = 35 u, O = 16 u)
Answer
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Hint: First of all write a balanced chemical reaction of decomposition of potassium chlorate ($KCl{{O}_{3}}$ ). Then for further calculations, use mole concept followed by unitary method so as to find the amount of potassium chlorate required for decomposition to obtain 9.6g of oxygen (${{O}_{2}}$).
Formula used:
We will require the following relationship:-
$w=n\times M$
where,
w = mass of the compound used or produced in the reaction.
n = number of moles.
M = Molar mass of the compound.
Complete answer:
As we know that decomposition reaction is defined as a reaction in which a compound breaks down into two or more simpler substances. The balanced chemical equation for the decomposition reaction of potassium chlorate is shown below:-
$2KCl{{O}_{3}}\to 2KCl+3{{O}_{2}}$
From the reaction, we can see that 2 moles of $KCl{{O}_{3}}$are involved in the reaction which means the mass of $KCl{{O}_{3}}$= $w=n\times M$
n = 2 moles
M = 39 + 35 + 3(16) = 122g/mol
$w=2moles\times 122g/mol$= 244g
From the reaction, we can see that 3 moles of ${{O}_{2}}$are produced in the reaction which means the mass of ${{O}_{2}}$= $w=n\times M$
n = 3 moles
M = 2(16) = 32g/mol
$w=3moles\times 32g/mol$= 96g
-Calculation of mass of potassium chlorate required for decomposition to obtain 9.6g of oxygen (${{O}_{2}}$):-
$2KCl{{O}_{3}}\to 2KCl+3{{O}_{2}}$
As we can see that 3 moles of ${{O}_{2}}$were produced from decomposition of 2 moles of $KCl{{O}_{3}}$which means 96g of ${{O}_{2}}$were produced from decomposition of 244g of $KCl{{O}_{3}}$.
Using unitary method:-
1g of ${{O}_{2}}$was produced from decomposition of = $\dfrac{244g}{96g}$ of $KCl{{O}_{3}}$
Hence, 9.6g of ${{O}_{2}}$was produced from decomposition of = $9.6g\times \dfrac{244g}{96g}=24.4g$ of $KCl{{O}_{3}}$
-Therefore, 24.4 grams of potassium chlorate should be decomposed to obtain 9.6 grams of oxygen.
Note: -
Always try to write balanced chemical reactions for such questions so as to know the number of moles of each compound involved in the reaction which is very helpful in further calculations of concentration terms.
-Kindly prefer to calculate the terms along with the units for better accuracy in results.
Formula used:
We will require the following relationship:-
$w=n\times M$
where,
w = mass of the compound used or produced in the reaction.
n = number of moles.
M = Molar mass of the compound.
Complete answer:
As we know that decomposition reaction is defined as a reaction in which a compound breaks down into two or more simpler substances. The balanced chemical equation for the decomposition reaction of potassium chlorate is shown below:-
$2KCl{{O}_{3}}\to 2KCl+3{{O}_{2}}$
From the reaction, we can see that 2 moles of $KCl{{O}_{3}}$are involved in the reaction which means the mass of $KCl{{O}_{3}}$= $w=n\times M$
n = 2 moles
M = 39 + 35 + 3(16) = 122g/mol
$w=2moles\times 122g/mol$= 244g
From the reaction, we can see that 3 moles of ${{O}_{2}}$are produced in the reaction which means the mass of ${{O}_{2}}$= $w=n\times M$
n = 3 moles
M = 2(16) = 32g/mol
$w=3moles\times 32g/mol$= 96g
-Calculation of mass of potassium chlorate required for decomposition to obtain 9.6g of oxygen (${{O}_{2}}$):-
$2KCl{{O}_{3}}\to 2KCl+3{{O}_{2}}$
As we can see that 3 moles of ${{O}_{2}}$were produced from decomposition of 2 moles of $KCl{{O}_{3}}$which means 96g of ${{O}_{2}}$were produced from decomposition of 244g of $KCl{{O}_{3}}$.
Using unitary method:-
1g of ${{O}_{2}}$was produced from decomposition of = $\dfrac{244g}{96g}$ of $KCl{{O}_{3}}$
Hence, 9.6g of ${{O}_{2}}$was produced from decomposition of = $9.6g\times \dfrac{244g}{96g}=24.4g$ of $KCl{{O}_{3}}$
-Therefore, 24.4 grams of potassium chlorate should be decomposed to obtain 9.6 grams of oxygen.
Note: -
Always try to write balanced chemical reactions for such questions so as to know the number of moles of each compound involved in the reaction which is very helpful in further calculations of concentration terms.
-Kindly prefer to calculate the terms along with the units for better accuracy in results.
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