What mass of hydrogen peroxide must decompose to produce 0.77 grams of water(H2O)?
Here is a worked example o
of a stoichiometry problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the mass of hydrogen peroxide needed to produce a given mass of water, we need to use the balanced chemical equation for the decomposition of hydrogen peroxide:
2 H2O2 ---> 2 H2O + O2
From this equation, we can see that 2 moles of hydrogen peroxide (H2O2) decompose to produce 2 moles of water (H2O).
To find the molar mass of water, we add the atomic masses of hydrogen (H) and oxygen (O):
H2O: (2 x 1.008 g/mol H) + (1 x 16.00 g/mol O) = 18.016 g/mol H2O
Therefore, 1 mole of water has a mass of 18.016 grams.
Now, let's calculate the number of moles of water produced from 0.77 grams using the following formula:
moles = mass / molar mass
moles of water = 0.77 g / 18.016 g/mol ≈ 0.0427 mol
Since the balanced equation tells us that 2 moles of hydrogen peroxide produce 2 moles of water, the moles of hydrogen peroxide needed to produce 0.0427 moles of water will also be 0.0427 moles.
Now, to find the mass of hydrogen peroxide, we multiply the number of moles by its molar mass. The molar mass of hydrogen peroxide (H2O2) is:
(2 x 1.008 g/mol H) + (2 x 16.00 g/mol O) = 34.014 g/mol H2O2
mass of H2O2 = moles of H2O2 × molar mass of H2O2
mass of H2O2 = 0.0427 mol × 34.014 g/mol H2O2 ≈ 1.4503 grams
Therefore, approximately 1.4503 grams of hydrogen peroxide must decompose to produce 0.77 grams of water.