The complete combustion of 0.492 , g of an or | vedclass

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

Question

(D) Moles of $CO_{2} = \frac{0.793}{44} \approx 0.01802 \, mol$. Since each $CO_{2}$ molecule contains one $C$ atom,moles of $C = 0.01802 \, mol$.Weig

Vedclass · Accepted Answer

$46$

Vedclass · Answer

(D) Moles of $CO_{2} = \frac{0.793}{44} \approx 0.01802 \, mol$. Since each $CO_{2}$ molecule contains one $C$ atom,moles of $C = 0.01802 \, mol$.Weight of $C = 0.01802 	imes 12 = 0.21624 \, g$.Moles of $H_{2}O = \frac{0.442}{18} \approx 0.02456 \, mol$. Since each $H_{2}O$ molecule contains two $H$ atoms,moles of $H = 0.02456 	imes 2 = 0.04912 \, mol$.Weight of $H = 0.04912 	imes 1 = 0.04912 \, g$.Weight of $O = 	ext{Total weight} - (	ext{Weight of } C + 	ext{Weight of } H)$Weight of $O = 0.492 - (0.21624 + 0.04912) = 0.492 - 0.26536 = 0.22664 \, g$.Percentage of $O = \frac{0.22664}{0.492} 	imes 100 \approx 46.06 \, \%$.Rounding to the nearest integer,we get $46 \, \%$.

The complete combustion of $0.492 \, g$ of an organic compound containing $'C'$,$'H'$,and $'O'$ gives $0.793 \, g$ of $CO_{2}$ and $0.442 \, g$ of $H_{2}O$. The percentage of oxygen composition in the organic compound is .... (nearest integer)

Similar Questions

On complete combustion,$0.492 \ g$ of an organic compound gave $0.792 \ g$ of $CO_2$. The $\%$ of carbon in the organic compound is $..................$ (Nearest integer)

The empirical formula of an acid is $CH_2O_2$. The probable molecular formula of the acid may be:

Actual number of atoms of different elements present in a molecule of a compound is given by

$A$ compound contains $2.2 \ g$ of phosphorus and $1.24 \ g$ of sulfur. What is its empirical formula?

What is the percentage of nitrogen in urea (in $\%$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module