When 3 g of a hydrocarbon is burnt in the pr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Moles of $CO_2 = \frac{8.8 \ g}{44 \ g/mol} = 0.2 \ mol$. $1 \ mol$ of $CO_2$ contains $12 \ g$ of $C$. So,$0.2 \ mol$ of $CO_2$ contains $0.2 	i

Vedclass · Accepted Answer

Law of Conservation of Mass

Vedclass · Answer

(A) Moles of $CO_2 = \frac{8.8 \ g}{44 \ g/mol} = 0.2 \ mol$. $1 \ mol$ of $CO_2$ contains $12 \ g$ of $C$. So,$0.2 \ mol$ of $CO_2$ contains $0.2 	imes 12 = 2.4 \ g$ of $C$. Moles of $H_2O = \frac{5.4 \ g}{18 \ g/mol} = 0.3 \ mol$. $1 \ mol$ of $H_2O$ contains $2 \ g$ of $H$. So,$0.3 \ mol$ of $H_2O$ contains $0.3 	imes 2 = 0.6 \ g$ of $H$. Total mass of $C$ and $H$ in the hydrocarbon $= 2.4 \ g + 0.6 \ g = 3.0 \ g$. Since the mass of reactants $(3.0 \ g)$ equals the mass of products $(2.4 \ g \ C + 0.6 \ g \ H = 3.0 \ g)$,this follows the Law of Conservation of Mass.

When $3 \ g$ of a hydrocarbon is burnt in the presence of sufficient oxygen,it produces $8.8 \ g$ of $CO_2$ and $5.4 \ g$ of $H_2O$. This process is an example of which law?

Similar Questions

Which of the following pairs of compounds cannot demonstrate the law of multiple proportions?

The statement related to the law of definite proportions is:

$1 \, L$ of $N_2$ combines with $3 \, L$ of $H_2$ to form $2 \, L$ of $NH_3$ under the same conditions. This illustrates the:

$n \ g$ of substance $X$ reacts with $m \ g$ of substance $Y$ to form $p \ g$ of substance $R$ and $q \ g$ of substance $S$. This reaction can be represented as,$X + Y \rightarrow R + S$. The relation which can be established in the amounts of the reactants and the products will be

Identify the symbol used for water according to $\text{Dalton's}$ atomic theory?

Vedclass Test Series

Exam Paper Generator

Online Exam Module