JEE MAIN - Chemistry (2019 - 12th April Evening Slot - No. 16)
25 g of an unknown hydrocarbon upon burning produces 88 g of CO2 and 9 g of H2O. This unknown
hydrocarbon contains :
18 g of carbon and 7 g of hydrogen
20 g of carbon and 5 g of hydrogen
22 g of carbon and 3 g of hydrogen
24 g of carbon and 1 g of hydrogen
Explanation
CxHy + $$\left( {x + {y \over 4}} \right)$$O2 $$ \to $$ xCO2 + $${y \over 2}$$H2O
No of moles of CO2 produced = $${{88} \over {44}}$$ = 2 mol
No of moles of H2O produced = $${9 \over {18}}$$ = 0.5 mol
According to stoichiometric mole mole analysis :
$${{moles\,of\,C{O_2}} \over x} = {{moles\,of\,{H_2}O} \over {{y \over 2}}}$$
$$ \Rightarrow $$ $${2 \over x} = {{0.5} \over {{y \over 2}}}$$
$$ \Rightarrow $$ $${x \over y} = {2 \over 1}$$
So, $${{no\,of\,atoms\,of\,'C'\,in\,{C_x}{H_y}} \over {no\,of\,atoms\,of\,'H'\,in\,{C_x}{H_y}}} = {2 \over 1}$$
$$ \therefore $$ $${{Mass\,of\,'C'\,in\,{C_x}{H_y}} \over {Mass\,of\,'H'\,in\,{C_x}{H_y}}} = {{2 \times 12} \over {1 \times 1}} = {{24} \over 1}$$
No of moles of CO2 produced = $${{88} \over {44}}$$ = 2 mol
No of moles of H2O produced = $${9 \over {18}}$$ = 0.5 mol
According to stoichiometric mole mole analysis :
$${{moles\,of\,C{O_2}} \over x} = {{moles\,of\,{H_2}O} \over {{y \over 2}}}$$
$$ \Rightarrow $$ $${2 \over x} = {{0.5} \over {{y \over 2}}}$$
$$ \Rightarrow $$ $${x \over y} = {2 \over 1}$$
So, $${{no\,of\,atoms\,of\,'C'\,in\,{C_x}{H_y}} \over {no\,of\,atoms\,of\,'H'\,in\,{C_x}{H_y}}} = {2 \over 1}$$
$$ \therefore $$ $${{Mass\,of\,'C'\,in\,{C_x}{H_y}} \over {Mass\,of\,'H'\,in\,{C_x}{H_y}}} = {{2 \times 12} \over {1 \times 1}} = {{24} \over 1}$$
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