The balanced chemical equation for the electrolysis of water is:
2H2O → 2H2 + O2
This equation shows that for every two moles of water that are electrolyzed, one mole of oxygen gas is produced. To solve this problem, we need to first convert the given mass of water (83.7 grams) to moles of water.
The molar mass of water (H2O) is:
2(1.008 g/mol H) + 15.999 g/mol O = 18.015 g/mol
So, 83.7 grams of water is equal to:
83.7 g / 18.015 g/mol = 4.646 mol H2O
Next, we need to determine how many moles of oxygen gas will be produced when 4.646 moles of water are electrolyzed. Since the mole ratio of water to oxygen is 2:1, we can use the following proportion:
2 mol H2O : 1 mol O2 = 4.646 mol H2O : x mol O2
Solving for x, we get:
x mol O2 = (1 mol O2 / 2 mol H2O) * 4.646 mol H2O = 2.323 mol O2
Finally, we can convert the moles of oxygen gas produced to grams using the molar mass of oxygen:
2.323 mol O2 * 32.00 g/mol O2 = 74.3 g O2
Therefore, 83.7 grams of water will produce 74.3 grams of oxygen gas by electrolysis.