From a laboratory process designed to separate water into hydrogen and oxygen gas, a student collected 11.9 g of hydrogen and 78.5 g of oxygen. How much water was originally involved in the process?

10.2g H * 1mol/1g (molar mass) = 10.2mol H

79.5g O * 1mol/16g = about 5 mol O
so about 5 moles of water (H2O so there should be twice as much H as O)
Trust me I'm sure this is right I do after all have an A+ in science so I'm positive I'm right

To determine the amount of water involved in the process, we need to use the concept of molar ratios. The molar ratio refers to the proportion of moles between different substances involved in a chemical reaction. In this case, we're interested in the molar ratio between water and hydrogen gas.

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

2H2O -> 2H2 + O2

According to this equation, for every 2 moles of water, we obtain 2 moles of hydrogen gas and 1 mole of oxygen gas.

First, let's calculate the number of moles of hydrogen gas collected:

Mass of hydrogen gas = 11.9 g
Molar mass of hydrogen gas (H2) = 2 g/mol (from the periodic table)

Number of moles of hydrogen gas = Mass of hydrogen gas / Molar mass of hydrogen gas
= 11.9 g / 2 g/mol
= 5.95 mol

Next, let's calculate the number of moles of oxygen gas collected:

Mass of oxygen gas = 78.5 g
Molar mass of oxygen gas (O2) = 32 g/mol (from the periodic table)

Number of moles of oxygen gas = Mass of oxygen gas / Molar mass of oxygen gas
= 78.5 g / 32 g/mol
= 2.45 mol

Since the molar ratio between water and hydrogen gas is 2:2 (from the balanced chemical equation), we can conclude that the number of moles of water is the same as the number of moles of hydrogen gas.

Therefore, the amount of water originally involved in the process is 5.95 moles.

To calculate the mass of water, we can use the molar mass of water, which is 18 g/mol.

Mass of water = Number of moles of water x Molar mass of water
= 5.95 mol x 18 g/mol
= 107.1 g

Hence, the original amount of water involved in the process was 107.1 grams.