Mass of candle 0.72 g before burning. What is the mass of the products of this reaction? Would that be the same thing as weighing the candle again (the mass is 0.27 g after burning)?

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Mass of candle 0.72 g before burning. What is the mass of the products of this reaction? Would that be the same thing as weighing the candle again (the mass is 0.27 g after burning)?

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Answer 1 · 2017-05-05T10:05:14

Refer to explanation

Explanation:

Reweighing the candle before and after burning just shows how much fuel was burnt.
Typical candle wax is an alkane such as hentriacontane, so write a chemical formula for the combustion of this alkane.

$C_{31} H_{64} + 47 O_{2} \to 31 C O_{2} + 32 H_{2} O$ $C_31H_64 + 47O_2 -> 31CO_2 + 32H_2O$

Work out the moles for this alkane.

$\frac{0.72 g}{436}$ $(0.72g)/(436)$ = 0.00165 moles

The mole ratio of the alkane to $C O_{2}$ $CO_2$ is 1:31 so multiply the moles of the alkane by 31 to get the number of moles for $C O_{2}$ $CO_2$ .

$0.00165 \cdot 31$ $0.00165*31$ =0.0511 moles

Multiply the moles of $C O_{2}$ $CO_2$ by $24 d m^{3}$ $24dm^3$ and then by 1000 to get it in $c m^{3}$ $cm^3$ .

$1000 (0.0511 \cdot 24) = 1266.4 c m^{3}$ $1000(0.0511*24) = 1266.4cm^3$

Hence for water, do 0.0511 moles times by its formula mass, 18.

#0.0511*18= 0.92g of water.

(I may have overcomplicated it a bit)

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Mass of candle 0.72 g before burning. What is the mass of the products of this reaction? Would that be the same thing as weighing the candle again (the mass is 0.27 g after burning)?

1 Answer

Explanation:

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