Question

The density of a wood is 0.79g/cm3. If the empirical formula of wood is CH2O. Calculate the mass of water produced when a log of dimension 12cm *14cm* 25cm is burnt completely

Answer 1

Answer on the Question #65735, Chemistry / Inorganic chemistry

The density of a wood is $0.79\mathrm{g/cm^3}$. If the empirical formula of wood is CH2O. Calculate the mass of water produced when a log of dimension $12\mathrm{cm} \times 14\mathrm{cm} \times 25\mathrm{cm}$ is burnt completely.

Solution:

Reaction of the wood combustion:

$$\mathrm{CH_2O} + \mathrm{O_2} = \mathrm{CO_2} + \mathrm{H_2O}$$

The mass of the wood is a composition of the density and volume of the piece of wood:

$$m(CH_2O) = d(CH_2O) \cdot V(CH_2O)$$ $$V(CH_2O) = a \cdot b \cdot c = 12\ \mathrm{cm} \cdot 14\ \mathrm{cm} \cdot 25\ \mathrm{cm} = 4200\ \mathrm{cm^3}$$ $$m(CH_2O) = 0.79\ \frac{g}{\mathrm{cm^3}} \cdot 4200\ \mathrm{cm^3} = 3318\ \mathrm{g}$$

The mole number of the wood equal to the mole number of the water (by the reaction):

$$n(CH_2O) = n(H_2O)$$ $$n(CH_2O) = \frac{m(CH_2O)}{M(CH_2O)} = \frac{3318\ \mathrm{g}}{30\ \mathrm{g/mol}} = 110,6\ \mathrm{mol}$$ $$m(H_2O) = n(H_2O) \cdot M(H_2O) = 110.6\ \mathrm{mol} \cdot 18\ \frac{\mathrm{g}}{\mathrm{mol}} = 1990.8\ \mathrm{g}$$

**Answer**: the 1990.8 g of the water produced.

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