Question #090a5

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Question #090a5

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Answer 1 · 2016-04-24T15:05:17

I don't think it can.

Explanation:

Since this is combustion, there must also by oxygen present which will contribute to the mass of the reactants.

However, if you consider the simplest organic compound, methane, combusting:
$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$ $CH_4+2O_2toCO_2+2H_2O$

The Relative Formula masses are:
$(16) methane + (2 \times 32) oxygen \to 44 (C O_{2}) + (2 \times 18) w a t e r$ $(16) "methane"+(2times32) "oxygen"to 44 (CO_2)+(2times18)water$

The increase in mass of products (80) to reactant (excluding oxygen) is a factor of 5.

The question requires a substance that gives this ratio to be
$\frac{3 + 1.5}{7.5 \times 10^{- 3}} = 600$ $(3+1.5)/(7.5times 10^-3)=600$
i.e. an increase in mass of a factor of 600.

This is not possible. If you consider the oxidation of individual atoms:
$4 H + O_{2} \to 2 H_{2} O$ $4H + O_2 to 2H_2O$ , the ratio by mass of water products to hydrogen reactants is 9.
For $C + O_{2} \to C O_{2}$ $C+O_2"toCO_2$ , the ratio by mass of $C O_{2}$ $CO_2$ products to carbon reactants is 3.7.

Hence we cannot get to a ratio of 600.

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Question #090a5

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