Question

0.22g of a hydrocarbon ( a compound containing carbon and hydrogen ) on combustion with 1.12g of oxygen gave 0.9g of water and 0.44g of carbon dioxide. show that these observations are accordance with the law of conservation of mass. ​

Answer 1

Answer:

(0.22+1.12)g = (0.90+0.44) g

Answer 2

These observations are accordance with the law of conservation of mass.

What is the Law of conservation of mass ?

• The Law of conservation of mass states that mass is conserved during a chemical change.
• The total mass of reactants is exactly equal to the total mass of the products in the reaction.
• For a chemical reaction $A+B\rightarrow C+D$, we will have  $mass_{reactant}=mass_{product}$
• So, $mass\hspace{1 mm}of\hspace{ 1 mm}A + mass\hspace{1 mm}of\hspace{ 1 mm}B= mass\hspace{1 mm}of\hspace{ 1 mm}C + mass\hspace{1 mm}of\hspace{ 1 mm}D$

We can see that for this question, we have a reaction like this:

$hydrocarbon + O_2 \rightarrow H_2O + CO_2$

Given:

• mass of hydrocarbon= 0.22 g
• mass of oxygen= 1.12 g
• mass of water= 0.9 g
• mass of carbon dioxide= 0.44 g

So, once we apply the law of conservation of mass:

$LHS=mass_{hydrocarbon}+mass_{oxygen}= (0.22 + 1.12) g= 1.34\hspace{1 mm} g$

$RHS=mass_{water}+mass_{CO_2}= (0.9 + 0.44) g= 1.34\hspace{1 mm} g$

We can see that $LHS=RHS$ which means that the Law of conservation of mass holds for this chemical reaction.

So, these observations are accordance with the law of conservation of mass.