Question

Calculate the weight of lime $CaO$ that can be obtained by heating 200Kg of limestone which is 93% pure. How many moles of impure potassium chlorate of 75% purity is required to produce 48g of oxygen?

Answer 1

Hint: $CaC{{O}_{3}}$ on heating produces Calcium oxide and carbon dioxide on heating. Potassium Chlorate on heating produces potassium chloride. Moles can be calculated dividing given mass by molecular mass. Chemical formula of Limestone is $CaC{{O}_{3}}$.

Complete step by step answer:
-one mole of Limestone on heating produces one mole of calcium oxide and one mole of oxygen.
\[CaC{{O}_{3}}\to CaO+{{O}_{2}}\]
-one mole is equivalent to Molecular mass.
Molecular weight of Calcium Carbonate is 100g.
Molecular weight of Calcium oxide is 56g
As 100g of calcium carbonate produces 56 g of Calcium oxide,
So 200Kg of Calcium Carbonate will produce 56Kg of Calcium Oxide.
As Limestone i.e. Calcium Carbonate is 93% pure ,
So Weight of \[CaO\] that can be obtained by heating 200kg of limestone
$=56000\times \dfrac{93}{100}=52080g$
So 52.08kg of \[CaO\] can be obtained by heating 200Kg of limestone..

-One mole of Potassium Chlorate On heating produces one mole of potassium chloride and 1.5 mole of Oxygen.
$KCl{{O}_{3}}\to KCl+\dfrac{3}{2}{{O}_{2}}$
`Molecular weight of Potassium Chlorate=122.5g
1 mole of oxygen gas =32g
So 1.5 mole of oxygen gas=48g
So , 122.5 g of Potassium Chlorate produces 48g of Oxygen.
As Potassium Chlorate is 75% pure, so amount of impure potassium Chlorate required to produce 48g of oxygen
\[=122.5\times \dfrac{100}{75}=163.3g\]
So 163.3g of impure Potassium Chlorate will be required to produce 48g of oxygen.

Note: 1 mole occupies $22.4d{{m}^{3}}$ of volume .one mole contains $6.022\times {{10}^{23}}$ molecules, ions or atoms. Mole concept can be used to find out the weight of a product or reactant produced in a reaction. % Purity is the percentage of pure compounds in an impure sample.