Question

On Combustion Of $25\mathrm{g}$ Of An Unknown Hydrocarbon Produces $88\mathrm{g}$ of ${\mathrm{CO}}_2$ and $9\mathrm{g}$ of ${\mathrm{H}}_2\mathrm{O}$. This Unknown Hydrocarbon Contains:

• A. $18\mathrm{g}$ of carbon and $7\mathrm{g}$ of hydrogen
• B. $20\mathrm{g}$ of carbon and $5\mathrm{g}$ of hydrogen
• C. $22\mathrm{g}$ of carbon and $3\mathrm{g}$ of hydrogen
• D. $24\mathrm{g}$ of carbon and $1\mathrm{g}$ of hydrogen

Answer 1

The correct option is D

$24\mathrm{g}$ of carbon and $1\mathrm{g}$ of hydrogen

The explanation for the Correct answer:

Option D $24\mathrm{g}$ of carbon a$1\mathrm{g}$ of hydrogen

Step 1: Amount of Carbon in $88\text{g}$ of $\text{C}{\text{O}}_{\text{2}}$

One mole of Carbon dioxide contains one mole of Carbon.

The given mass of Carbon dioxide is $88\text{g}$

The molar masses of Carbon dioxide is $44\mathrm{g}$

The mass of Carbon is $12\mathrm{g}$
Therefore, the amount of Carbon in $88\text{g}$ of $\text{C}{\text{O}}_{\text{2}}$$\mathrm{=88}\mathrm{gC}{\mathrm{O}}_2\times \frac{12\mathrm{g}C}{44\mathrm{g}{\mathrm{CO}}_2}=24\mathrm{g}$
Thus, the amount of Carbon in $88\text{g}$ of $\text{C}{\text{O}}_{\text{2}}$ is $24\mathrm{g}$.

Step 2: Amount of Hydrogen in $9\text{g}$ of ${\text{H}}_{\text{2}}\text{O}$

From the molecular formula of ${\text{H}}_{\text{2}}\text{O}$,

One mole of water contains two moles of Hydrogen.

The given mass of water is $9\mathrm{g}$

The molar masses of water is $18\mathrm{g}$

The mass of Hydrogen is $1\mathrm{g}$

Since water contains 2 moles of hydrogen, the mass is of Hydrogen in water is $2\times 1=2\mathrm{g}$

Therefore, the amount of Hydrogen in $9\text{g}$ of ${\text{H}}_{\text{2}}\text{O}$ $=9\mathrm{g}{\mathrm{H}}_2\mathrm{O}\times \frac{2\mathrm{g}H}{18\mathrm{g}{\mathrm{H}}_2\mathrm{O}}=1\mathrm{g}$

The amount of hydrogen in $9\mathrm{g}$ of ${\text{H}}_{\text{2}}\text{O}$ is $1\mathrm{g}$

Therefore, The unknown hydrocarbon contains $24\text{g}$of Carbon and $1\text{g}$ of Hydrogen, and option D is correct.