Final answer:
The theoretical yield of water formed from the reaction of 64.6g of hexane and an excess of oxygen is 95.0 grams.
Step-by-step explanation:
To determine the theoretical yield of water (H₂O) from the reaction of liquid hexane (CH₃(CH₂)₄CH₃) with gaseous oxygen (O₂), we must first write a balanced chemical equation:
2 C₆H₁₄ + 19 O₂ → 12 CO₂ + 14 H₂O
From this balanced equation, we see that two moles of hexane produce 14 moles of water. Using the molar mass of hexane (86.18 g/mol) we can find out how many moles of hexane 64.6g corresponds to:
Moles of hexane = mass (g) / molar mass (g/mol) = 64.6 g / 86.18 g/mol = 0.75 moles (to three significant figures).
Since two moles of hexane give 14 moles of water, 0.75 moles will give:
0.75 moles hexane x (7 moles H₂O / 1 mole hexane) = 5.25 moles H₂O
The molar mass of water H₂O is 18.02 g/mol. The theoretical yield of water in grams would be:
Theoretical yield = moles H₂O x molar mass H₂O = 5.25 moles x 18.02 g/mol
= 94.605 g H₂O (95.0 g to three significant figures).
Therefore, the theoretical yield of water formed from 64.6g of hexane and an excess of oxygen is 95.0 grams.