the complete electrolysis of 90 grams of water will yield

To determine the products of the complete electrolysis of water, we need to consider the chemical equation for the electrolysis of water. The balanced equation for the electrolysis of water is:

2H2O(l) → 2H2(g) + O2(g)

From this equation, we can see that water decomposes into hydrogen gas (H2) and oxygen gas (O2) during electrolysis.

To find the amount of hydrogen gas and oxygen gas produced, we need to convert the mass of water (90 grams) to moles and then use the stoichiometry of the balanced equation.

Step 1: Convert mass of water to moles
Molar mass of water (H2O) = 2(1.00784 g/mol) + 15.999 g/mol ≈ 18.015 g/mol

moles of water = mass of water / molar mass of water
moles of water = 90 g / 18.015 g/mol ≈ 4.997 moles

Step 2: Use stoichiometry to find moles of hydrogen gas and oxygen gas
From the balanced equation, we can see that 2 moles of water produce 2 moles of hydrogen gas and 1 mole of oxygen gas.

moles of hydrogen gas = moles of water
moles of hydrogen gas = 4.997 moles

moles of oxygen gas = 0.5 * moles of water
moles of oxygen gas = 0.5 * 4.997 moles ≈ 2.498 moles

Step 3: Convert moles to grams
Molar mass of hydrogen gas (H2) = 2(1.00784 g/mol) ≈ 2.015 g/mol
Molar mass of oxygen gas (O2) = 2(15.999 g/mol) ≈ 31.998 g/mol

mass of hydrogen gas = moles of hydrogen gas * molar mass of hydrogen gas
mass of hydrogen gas = 4.997 moles * 2.015 g/mol ≈ 10.066 g

mass of oxygen gas = moles of oxygen gas * molar mass of oxygen gas
mass of oxygen gas = 2.498 moles * 31.998 g/mol ≈ 79.936 g

Therefore, the complete electrolysis of 90 grams of water will yield approximately 10.066 grams of hydrogen gas and 79.936 grams of oxygen gas.