Chemistry Stoichiometry: How Many Grams of Hydrogen Peroxide Needed to Produce Water?

Question:

What is the amount of hydrogen peroxide required to produce 15.6 g of water?

Answer:

To produce 15.6 g of water from hydrogen peroxide, approximately 29.48 g of hydrogen peroxide would be needed, as the reaction has a 1:1 molar ratio.

Chemistry Stoichiometry: Stoichiometry is a branch of chemistry that deals with the calculation of the quantities of reactants and products in chemical reactions.

In this particular case, the question asks about the amount of hydrogen peroxide (H₂O₂) needed to produce water (H₂O). The balanced chemical reaction for the decomposition of hydrogen peroxide is 2H₂O₂ → 2H₂O + O₂.

From the balanced equation, we can see that two moles of hydrogen peroxide produce two moles of water. This means there is a 1:1 molar ratio of H₂O₂ to H₂O in this reaction.

The molar mass of water is about 18 g/mol, so 15.6 g of water is approximately 0.867 moles. To find the necessary mass of hydrogen peroxide, we use its molar mass (34 g/mol) and the same amount of moles as in the case of water. Multiplying 0.867 moles by 34 g/mol gives us around 29.48 grams of hydrogen peroxide required to produce 15.6 g of water.

Stoichiometry is essential in determining the correct amounts of reactants needed for a chemical reaction and is based on the principles of conservation of mass and molar ratios.