
3g of hydrocarbon on combustion in the excess of oxygen produces 8.8 g of $ C{O_2} $ and 5.4 g of $ {H_2}O $ . The data illustrates the law of:
A. Conservation of mass
B. Multiple proportions
C. Constant proportions
D. Reciprocal proportion
Answer
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Hint: The law of chemical combination is the combination of five basic laws that govern the chemical combination of elements, these laws are:
Conservation of mass
Multiple proportions
Constant proportions
Reciprocal proportion
Gay Lussac’s law of gaseous volume
Complete step by step answer:
Here it is given that three grams of hydrocarbon on combustion with an excess of oxygen produced 8.8 g of carbon dioxide and 5.4 g of water and we know hydrocarbons are those compounds which contain hydrogen and carbon atoms. Also one mole of carbon dioxide contains one mole of a carbon atom. So the mass of one mole of carbon dioxide is calculated as:
Mass of carbon atom + 2\[ \times \]mass of oxygen atom
As we know, the mass of a carbon atom is 12 and the oxygen is 16. Hence, the mass of carbon dioxide is = $ 12 + 2 \times 16 = 44g $ . Similarly the weight of 8.8g of carbon dioxide atom is calculated as below:
Thus 44 g of carbon dioxide contains = 12 gram of carbon
1 g of carbon dioxide contain= $ \dfrac{{12}}{{44}}g $ of carbon
8.8 g of carbon dioxide contain= $ \dfrac{{12}}{{44}} \times 8.8 = 2.4g $
Similarly, the mass of 1 mole of water can be calculated as: $ $
$ 2 \times {\text{mass of hydrogen + mass of oxygen}} $
The mass of hydrogen is 1 and the mass of oxygen is 16 thus the mass of water is= $ 2 \times 1 + 16 = 18g $ . So to calculate the mass of 5.4g of water, the calculation is as given below:
18g of water contain=2 g of hydrogen
1g of water contain= $ \dfrac{2}{8}g $
5.4g of water contain= $ \dfrac{2}{8} \times 5.4 = 0.6g $
The total mass of hydrogen and carbon is = $ 2.4 + 0.6 = 3g $ , which is equal to the given mass of hydrocarbon. Hence we can say this illustrated the law of conservation of mass.
So, Option A is correct.
Note: The law of conservation of mass states that during the physical and chemical change the mass of the reactants is equal to the mass of products. Or it can be stated as the mass can neither be created nor destroyed.
Conservation of mass
Multiple proportions
Constant proportions
Reciprocal proportion
Gay Lussac’s law of gaseous volume
Complete step by step answer:
Here it is given that three grams of hydrocarbon on combustion with an excess of oxygen produced 8.8 g of carbon dioxide and 5.4 g of water and we know hydrocarbons are those compounds which contain hydrogen and carbon atoms. Also one mole of carbon dioxide contains one mole of a carbon atom. So the mass of one mole of carbon dioxide is calculated as:
Mass of carbon atom + 2\[ \times \]mass of oxygen atom
As we know, the mass of a carbon atom is 12 and the oxygen is 16. Hence, the mass of carbon dioxide is = $ 12 + 2 \times 16 = 44g $ . Similarly the weight of 8.8g of carbon dioxide atom is calculated as below:
Thus 44 g of carbon dioxide contains = 12 gram of carbon
1 g of carbon dioxide contain= $ \dfrac{{12}}{{44}}g $ of carbon
8.8 g of carbon dioxide contain= $ \dfrac{{12}}{{44}} \times 8.8 = 2.4g $
Similarly, the mass of 1 mole of water can be calculated as: $ $
$ 2 \times {\text{mass of hydrogen + mass of oxygen}} $
The mass of hydrogen is 1 and the mass of oxygen is 16 thus the mass of water is= $ 2 \times 1 + 16 = 18g $ . So to calculate the mass of 5.4g of water, the calculation is as given below:
18g of water contain=2 g of hydrogen
1g of water contain= $ \dfrac{2}{8}g $
5.4g of water contain= $ \dfrac{2}{8} \times 5.4 = 0.6g $
The total mass of hydrogen and carbon is = $ 2.4 + 0.6 = 3g $ , which is equal to the given mass of hydrocarbon. Hence we can say this illustrated the law of conservation of mass.
So, Option A is correct.
Note: The law of conservation of mass states that during the physical and chemical change the mass of the reactants is equal to the mass of products. Or it can be stated as the mass can neither be created nor destroyed.
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