72.1 g of steam is reacted with excess ethene (C2H4) according to the following reaction.                                    C2H4(g) + H2O(g) → C2H5OH(l) Determine the maximum mass (in grams) of ethanol that could be produced. Give the answer to 3 significant figures.

To solve this problem, we need to use stoichiometry, which is a method in chemistry that uses balanced chemical equations to calculate quantities of reactants or products.

1. First, we need to find the molar mass of water (H2O). The molar mass of hydrogen (H) is approximately 1 g/mol and the molar mass of oxygen (O) is approximately 16 g/mol. Therefore, the molar mass of water is (2*1) + 16 = 18 g/mol.

2. Next, we need to convert the mass of steam (water in gaseous form) to moles. We do this by dividing the given mass of steam by the molar mass of water. So, 72.1 g ÷ 18 g/mol = 4.005 mol of H2O.

3. The balanced chemical equation shows that one mole of ethene (C2H4) reacts with one mole of water (H2O) to produce one mole of ethanol (C2H5OH). Therefore, the number of moles of ethanol that can be produced is the same as the number of moles of water, which is 4.005 mol.

4. Finally, we need to convert the moles of ethanol to grams. The molar mass of ethanol is approximately 46 g/mol (212 for carbon, 61 for hydrogen, and 16 for oxygen). Therefore, the maximum mass of ethanol that can be produced is 4.005 mol * 46 g/mol = 184.23 g.

So, the maximum mass of ethanol that could be produced from 72.1 g of steam is approximately 184 g (to three significant figures).