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  • If a rocket runs on fuel <img alt="\mathrm{\left ( C_{15}H_{30} \right )}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cleft%20%28%20C_%7B15%7DH_%7B30%7D%20%5Cright%20%29%7D"

If a rocket runs on fuel \mathrm{\left ( C_{15}H_{30} \right )}  and liquid oxygen,the weight of oxygen required and \mathrm{CO_{2}} released for every litre of fuel respectively are:
(Given: density of the fuel is \mathrm{0.756\, g/mL})

Option: 1

\mathrm{1188\, g\, and\, 1296\, g}


Option: 2

\mathrm{2376\, g\, \,and\,\, 2592\, g}


Option: 3

\mathrm{2592\, g\, \,and\,\, 2376\, g}


Option: 4

\mathrm{3429\, g\, \,and\,\, 3142\, g}


Answers (1)

best_answer

The combustion reaction is given as
\mathrm{C}_{15}\, \mathrm{H}_{30}+\frac{45}{2} \mathrm{O}_{2}\, \rightarrow \, 15\, \mathrm{CO}_{2}+15\, \mathrm{H}_{2} \mathrm{O}
So,
1 \text { mole fuel } \equiv \frac{45}{2} \text { moles of } \mathrm{O}_{2} \equiv 15 \text { mole of } \mathrm{CO}_{2}
now,
1 lit of fuel weights  \mathrm{756\, g}
\mathrm{\therefore\: \text{moles in 1 lit}\: of \: fuel =\frac{756}{210}=3.6}

\mathrm{\therefore \: moles\: of \: \mathrm{O}_{2}=3.6 \times \frac{45}{2}=81}
     \mathrm{and \: moles\: of \: \mathrm{CO}_{2}=3.6 \times 15=54}

\mathrm{\therefore\: weight \: of \: \mathrm{O}_{2}=81 \times 32=2592 \mathrm{~g}}
      \mathrm{weight \: of \: \mathrm{CO}_{2}=337.5 \times 44=2376 \mathrm{~g}}

Hence, the correct answer is Option (3)

Posted by

Anam Khan

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