Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . suppose 41. g of hexane is mixed with 150. g of oxygen. calculate the maximum mass of water that could be produced by the chemical reaction. round your answer to significant digits.

The balanced chemical equation for the reaction of hexane with O₂ is given below;
2C₆H₁₄  + 19O₂ --> 12CO₂ + 14H₂O
Stoichiometry of hexane to oxygen is 2:19
number of hexane moles - 41 g / 86.2 g/mol = 0.476 mol
number of oxygen moles - 150 g / 32 g/mol = 4.69 mol
we need to find which reagent is the limiting reactant 
If hexane is the limiting reactant 
2 moles of hexane reacts with -19 mol of oxygen 
therefore 0.476 mol - 19/2 x 0.476 = 4.522 mol
since 4.69 mol of oxygen is present, O₂ is in excess and hexane is the limiting reactant.
stoichiometry of hexane to water is 2:14
2 mol of hexane gives 14 mol of water
therefore 0.476 mol of hexane forms - 14/2 x 0.476 - 3.33 mol
mass of water formed - 3.33 mol x 18.0 g/mol = 59.9 g

So what you would do is go through and add 3 teaspoons(tsp) 6 times because of the 6 tablespoons(tbsp). which will give you an answer of 18. another way is to multiply 3x6=18.