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How much water (H2O) reacts to form 3.0 g of Hydrogen gas and 24.0 g of oxygen gas during the þrocess of electrolysis?​

User Thoomas
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Final answer:

36.04 grams of water reacts to form 3.0 g of hydrogen gas and 24.0 g of oxygen gas during electrolysis.

Step-by-step explanation:

The balanced equation for the reaction of water (H2O) to form hydrogen gas (H2) and oxygen gas (O2) during electrolysis is:

2 H2O(l) → 2 H2(g) + O2(g)

To determine how much water reacts to form 3.0 g of hydrogen gas and 24.0 g of oxygen gas, we need to calculate the amount of water in grams using the molar masses of hydrogen and oxygen:

2 moles of hydrogen gas (H2) = 2 x 2.02 g = 4.04 g

1 mole of oxygen gas (O2) = 32.0 g

Therefore, the total mass of water (H2O) that reacts is: 4.04 g + 32.0 g = 36.04 g

So, 36.04 grams of water reacts to form 3.0 g of hydrogen gas and 24.0 g of oxygen gas during electrolysis.

User Kshitij Mittal
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