To calculate the grams of ethylene required to react with 0.132 mol of water (H2O), we need to know the balanced chemical equation for the reaction between ethylene (C2H4) and water.

The balanced equation is:

C2H4(g) + H2O(g) → C2H5OH(g)

From this equation, we can see that 1 mol of ethylene reacts with 1 mol of water.

Therefore, the moles of ethylene required are equal to the moles of water:

Moles of ethylene = Moles of water = 0.132 mol

Now, we can convert moles of ethylene to grams using its molar mass:

Molar mass of ethylene (C2H4) = 2(12.01 g/mol) + 4(1.008 g/mol) = 28.05 g/mol

Grams of ethylene = Moles of ethylene × Molar mass of ethylene

Grams of ethylene = 0.132 mol × 28.05 g/mol

Grams of ethylene = 3.70 grams (approximately)

Therefore, 3.70 grams of ethylene are needed to react with 0.132 mol of water.