Calculate the mass of ethanol produced from 1.0 kg of ethene according to the reaction C2H4 + H2O -> C2H5OH.

The relative atomic masses are $\text{A}_r$: C = 12, H = 1, O = 16.
Molar mass of ethene ($\text{C}_2\text{H}_4$):

$$M(\text{C}_2\text{H}_4) = (2 \times 12) + (4 \times 1) = 24 + 4 = 28 \text{ g/mol}$$

Molar mass of ethanol ($\text{C}_2\text{H}_5\text{OH}$):

$$M(\text{C}_2\text{H}_5\text{OH}) = (2 \times 12) + (5 \times 1) + (1 \times 16) + (1 \times 1) = 24 + 5 + 16 + 1 = 46 \text{ g/mol}$$

Given mass of ethene = $1.0 \text{ kg} = 1000 \text{ g}$.

$$\text{Moles of ethene} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{1000 \text{ g}}{28 \text{ g/mol}}$$ $$\text{Moles of ethene} \approx 35.714 \text{ mol}$$

The balanced chemical equation is $\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}$.
From the equation, 1 mole of ethene produces 1 mole of ethanol.
Therefore, moles of ethanol produced = moles of ethene.

$$\text{Moles of ethanol} = \frac{1000}{28} \text{ mol}$$

$$\text{Mass of ethanol} = \text{Moles of ethanol} \times \text{Molar mass of ethanol}$$ $$\text{Mass of ethanol} = \left(\frac{1000}{28} \text{ mol}\right) \times (46 \text{ g/mol})$$ $$\text{Mass of ethanol} = \frac{1000 \times 46}{28} \text{ g}$$ $$\text{Mass of ethanol} = \frac{46000}{28} \text{ g}$$ $$\text{Mass of ethanol} \approx 1642.857 \text{ g}$$

Convert the mass to kilograms:

$$\text{Mass of ethanol} = \frac{1642.857}{1000} \text{ kg} \approx 1.642857 \text{ kg}$$

Rounding to three significant figures:

$$\text{Mass of ethanol} = \boxed{\text{1.64 kg}}$$