If 25.4 grams of water react to form 2.8 grams of hydrogen, how many grams of oxygen must simultaneously be formed?

There will be formed 22.6 grams O2

Explanation:

Step 1: Data given

Mass of 25.4 grams

Mass of hydrogen formed = 2.8 grams

Step 2: The balanced equation

2H2O (l) → 2H2 (aq) + O2 (aq)

For 2 moles H2O we'll have 2 moles hydrogen and 1 mol O2

Step 3: Calculate moles of H2O

Moles H2O = mass H2O / molar mass H2O

Moles H2O = 25.4 grmas / 18.02 g/mol

Moles H2O = 1.41 moles

Step 4: Calculate moles hydrogen formed

Moles H2 = 2.8 grams / 2.0 g/mol

Moles H2 = 1.4 moles

Step 5: Calculate moles of O2

For 2 mole of H2O reacted, there will be formed 1 mol of H2O

For 1.41 moles of H2O formed there will be formed 0.705moles O2.

Step 6: Calculate mass O2

Mass O2 = moles O2 * molar mass O2

Mass O2 = 0.705 * 32 g/mol

Mass O2 = 22.6 grams O2

There will be formed 22.6 grams O2

The mass of oxygen and hydrogen must be equal to the mass of the substance they create the water. So if the hydrogen is 2.8 g the oxygen must account for the rest of the mass. Basically just subtract 25.4-2.8=mass of oxygen