From a laboratory process designed to separate water into hydrogen and oxygen gas, a student collected 10.0 g of hydrogen and 70.9 g of oxygen. How much water was originally involved in the process?

80.9 grams it change forms but not values

To find out how much water was originally involved in the process, we need to use the law of conservation of mass. According to this law, mass can neither be created nor destroyed during a chemical reaction.

The chemical equation for the separation of water into hydrogen and oxygen gas is:

2H₂O → 2H₂ + O₂

From the balanced equation, we can see that for every 2 moles of water, we obtain 2 moles of hydrogen gas and 1 mole of oxygen gas. The molar mass of water (H₂O) is approximately 18 g/mol, the molar mass of hydrogen (H₂) is approximately 2 g/mol, and the molar mass of oxygen (O₂) is approximately 32 g/mol.

Using the given mass of hydrogen (10.0 g) and the molar mass of hydrogen (2 g/mol), we can calculate the number of moles of hydrogen:

Number of moles of hydrogen = Mass of hydrogen / Molar mass of hydrogen
Number of moles of hydrogen = 10.0 g / 2 g/mol
Number of moles of hydrogen = 5 mol

Since we have 2 moles of hydrogen for every 2 moles of water, this means that the number of moles of water involved in the process is also 5 mol.

Now, using the number of moles of water (5 mol) and the molar mass of water (18 g/mol), we can calculate the mass of water:

Mass of water = Number of moles of water × Molar mass of water
Mass of water = 5 mol × 18 g/mol
Mass of water = 90 g

Therefore, the original amount of water involved in the process was 90 grams.