what is the amount of water produced when 8 g of hydrogen is reacted with 32 g of oxygen

The amount of water produced is 36 g.

Quick Scoop

To answer “what is the amount of water produced when 8 g of hydrogen is reacted with 32 g of oxygen,” we use the balanced equation for water formation.

1. Balanced reaction

The reaction between hydrogen and oxygen to form water is:

2H2+O2→2H2O2H_2+O_2\rightarrow 2H_2O2H2​+O2​→2H2​O

This tells us that:

• 2 moles of H2H_2H2​ react with 1 mole of O2O_2O2​
• to produce 2 moles of H2OH_2OH2​O.

2. Moles of reactants

• Molar mass of H2H_2H2​ = 2 g/mol.

* Moles of H2H_2H2​ in 8 g = 8÷2=48\div 2=48÷2=4 moles.

• Molar mass of O2O_2O2​ = 32 g/mol.

* Moles of O2O_2O2​ in 32 g = 32÷32=132\div 32=132÷32=1 mole.

From the stoichiometric ratio (2 mol H2H_2H2​ : 1 mol O2O_2O2​),

• 4 moles of H2H_2H2​ would need 2 moles of O2O_2O2​.

• But only 1 mole of O2O_2O2​ is present, so oxygen is the limiting reagent.

3. Water produced

According to the balanced equation:

• 1 mole of O2O_2O2​ produces 2 moles of H2OH_2OH2​O.

So:

• Moles of H2OH_2OH2​O formed = 2 moles.

• Molar mass of H2OH_2OH2​O = 18 g/mol.

• Mass of water = 2×18=362\times 18=362×18=36 g.

Final result: When 8 g of hydrogen reacts with 32 g of oxygen, 36 g of water is produced, with hydrogen left in excess and oxygen completely consumed.

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Find out the amount of water formed when 8 g of hydrogen reacts with 32 g of oxygen, using the balanced equation 2H2+O2→2H2O2H_2+O_2\rightarrow 2H_2O2H2​+O2​→2H2​O, limiting reagent, and stoichiometry.