The balanced equation for the synthesis of water is:

2H2 + O2 > 2H2O

If I have 5.5 grams of water, at least how many grams of Oxygen gas was needed?

I believe the answer is 2.75grams of Oxygen Gas

H2O = 18 g/mol

so I have 5.5/18 mols water

you get two mols H2O for every mol of O2

so I used 5.5/36 mols O2
O2 = 32 g/mol

so
5.5*32/36 = 4.89 grams O2

To determine how many grams of oxygen gas was needed to produce 5.5 grams of water, we need to use the balanced equation for the synthesis of water.

The balanced equation is:

2H2 + O2 > 2H2O

From the equation, we can see that for every 2 moles of hydrogen gas (H2) consumed, 1 mole of oxygen gas (O2) is consumed, and 2 moles of water (H2O) are produced.

To find the number of moles of water, we divide the mass of water (5.5 grams) by its molar mass. The molar mass of water (H2O) is approximately 18 g/mol.

Number of moles of water = 5.5 g / 18 g/mol

Next, we need to compare the mole ratio between oxygen and water in the balanced equation. According to the equation, 2 moles of water are produced for every mole of oxygen gas consumed.

So, the number of moles of oxygen gas needed would be half of the number of moles of water:

Number of moles of oxygen gas = (5.5 g / 18 g/mol) / 2

Now we can calculate the mass of oxygen gas by multiplying the number of moles by its molar mass. The molar mass of oxygen (O2) is approximately 32 g/mol.

Mass of oxygen gas = ( (5.5 g / 18 g/mol) / 2 ) * 32 g/mol

By evaluating this expression, we can find the answer to your question.