Write the balanced equation when 9 grams of water undergo electrolysis and calculate the amount of oxygen gas liberated?
4 grams
I don't think so. Are you guessing? Post your work and I'll find the error.
To write the balanced equation for the electrolysis of water, we need to understand that water molecules (H2O) can be broken down into hydrogen gas (H2) and oxygen gas (O2) through the process of electrolysis.
The balanced equation for the electrolysis of water is:
2H2O(l) → 2H2(g) + O2(g)
This equation shows that two water molecules (H2O) can be broken down into two molecules of hydrogen gas (H2) and one molecule of oxygen gas (O2).
Now, to calculate the amount of oxygen gas liberated, we need to use the concept of mole-to-mole ratios.
1 mole of water (H2O) is equal to 18 grams (mass of H2O).
From the given information of 9 grams of water, we can calculate the number of moles of water involved in the reaction:
9 grams H2O × (1 mole H2O / 18 grams H2O) = 0.5 moles H2O
According to the balanced equation, the mole ratio between water and oxygen gas is 1:1. Therefore, the number of moles of oxygen gas produced is also 0.5 moles.
To convert moles into grams, we can use the molar mass of oxygen gas, which is approximately 32 grams per mole.
0.5 moles O2 × (32 grams O2 / 1 mole O2) = 16 grams O2
Therefore, 9 grams of water undergoing electrolysis would result in the liberation of approximately 16 grams of oxygen gas.