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

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(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 $2376 \ g$

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(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 $0.756 \ g/mL$)

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