If a rocket runs on a fuel (C_{15}H_{30}) and | vedclass

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

Question

(C) The balanced chemical equation for the combustion of the fuel is:$C_{15}H_{30} + \frac{45}{2} O_{2} \rightarrow 15 CO_{2} + 15 H_{2}O$First,calcul

Vedclass · Accepted Answer

$2592 \ g$ and $2376 \ g$

Vedclass · Answer

(C) The balanced chemical equation for the combustion of the fuel is:$C_{15}H_{30} + \frac{45}{2} O_{2} ightarrow 15 CO_{2} + 15 H_{2}O$First,calculate the mass of $1 \ L$ $(1000 \ mL)$ of fuel:$	ext{Mass} = 	ext{density} 	imes 	ext{volume} = 0.756 \ g/mL 	imes 1000 \ mL = 756 \ g$The molar mass of $C_{15}H_{30}$ is $(15 	imes 12) + (30 	imes 1) = 210 \ g/mol$.Number of moles of fuel $= \frac{756 \ g}{210 \ g/mol} = 3.6 \ mol$.According to the stoichiometry,$1 \ mol$ of fuel requires $\frac{45}{2} = 22.5 \ mol$ of $O_{2}$.So,$3.6 \ mol$ of fuel requires $3.6 	imes 22.5 = 81 \ mol$ of $O_{2}$.Weight of $O_{2} = 81 \ mol 	imes 32 \ g/mol = 2592 \ g$.According to the stoichiometry,$1 \ mol$ of fuel produces $15 \ mol$ of $CO_{2}$.So,$3.6 \ mol$ of fuel produces $3.6 	imes 15 = 54 \ mol$ of $CO_{2}$.Weight of $CO_{2} = 54 \ mol 	imes 44 \ g/mol = 2376 \ g$.

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

Similar Questions

The moles of methane required to produce $81 \ g$ of water after complete combustion is $.... \times 10^{-2} \ mol.$ [nearest integer]

$x$ grams of calcium carbonate was completely burnt in air. The weight of the solid residue formed is $28 \,g$. What is the value of $x$ (in grams)?

$1 \text{ mole}$ of concentrated $HCl$ requires $X \text{ moles}$ of dilute $NaOH$ for neutralization,and $1 \text{ mole}$ of concentrated $H_2SO_4$ requires $Y \text{ moles}$ of dilute $NaOH$. Which of the following relations is true?

For the given series of reactions:
$NH_{3(g)} + O_{2(g)} \rightarrow NO_{(g)} + H_2O_{(l)}$
$NO_{(g)} + O_{2(g)} \rightarrow NO_{2(g)}$
To obtain the maximum mass of $NO_2$ from a given mass of a mixture of $NH_3$ and $O_2$,the ratio of the mass of $NH_3$ to $O_2$ should be:

Six volumes of oxygen,on complete ozonisation,form ...... volumes of ozone.

Vedclass Test Series

Exam Paper Generator

Online Exam Module