The alcohol in gasohol burns according to the reaction

C2H5OH(l) + 3O2(g) --> 2CO2(g) + 3H2O

how many grams of water are produced when 3.0 moles of C2H5OH are burned in an excess of oxygen gas?

mols C2H5OH = 3.0

mols H2O produced = 3.0 mols EtOH x (3 mols H2O/1 mol EtOH) = 9 mols H2O produced.
g H2O = mols x molar mass.

To find the number of grams of water produced when 3.0 moles of C2H5OH are burned, we will use the balanced chemical equation provided.

The balanced chemical equation is:

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O

From the equation, we can see that for every 1 mole of C2H5OH burned, 3 moles of water are produced. Therefore, we can set up a mole-to-mole ratio:

3.0 moles C2H5OH * (3 moles H2O / 1 mole C2H5OH)

Now we can calculate the number of moles of water produced:

3.0 moles C2H5OH * (3 moles H2O / 1 mole C2H5OH) = 9.0 moles H2O

Finally, we can convert moles of water to grams using the molar mass of water. The molar mass of water (H2O) is approximately 18.0153 g/mol:

9.0 moles H2O * 18.0153 g/mol = 162.14 grams of H2O

Therefore, when 3.0 moles of C2H5OH are burned in an excess of oxygen gas, approximately 162.14 grams of water are produced.