How much oxygen gas can I make from a #9*kg# mass of water?

Question

Mateo Aguilar · Accepted Answer

#H_2O(g) rarr 1/2O_2(g) + H_2(g)# You have #90xx10^3*g# of water. This represents a MOLAR quantity of #(90xx10^3*g)/(18*g*mol^-1)=??mol#. The number of moles of oxygen gas, #O_2#, is half this molar quantity by the stoichiometric equation. And of course each mole of oxygen gas has a mass of #32*g#. From where am I getting these masses? Do you have to learn them?

Natalie Anton · Answer

undefined To calculate the amount of oxygen gas produced from a given mass of water, you can use the stoichiometry of the electrolysis of water, which produces hydrogen gas and oxygen gas in a 2:1 molar ratio.

1 mole of water (H2O) produces 1 mole of oxygen gas (O2) and 2 moles of hydrogen gas (H2).

Given that the molar mass of water is approximately 18 g/mol, and the molar mass of oxygen gas is approximately 32 g/mol:

1 mole of water = 2 moles of hydrogen gas = 1 mole of oxygen gas = 18 g

Using the given mass of water (9 kg or 9000 g), you can calculate the number of moles of water:

9000 g ÷ 18 g/mol = 500 moles of water

Since the ratio of water to oxygen gas is 1:1, the number of moles of oxygen gas produced is also 500 moles.

To convert moles of oxygen gas to grams, you multiply by the molar mass of oxygen gas:

500 moles × 32 g/mol = 16000 g

Therefore, you can produce approximately 16,000 grams or 16 kilograms of oxygen gas from a 9-kilogram mass of water.