Hydrogen Peroxide decomposes to produce oxygen gas and water, What mass of hydrogen peroxide (H2O2) nust decompose to produce 0.77 grams of water?

1. Write a balanced equation :

2H^2O^2---O^2+ 2H^2O
2. Find the molar mass (g/mol) of each:
34.02 32.00 18.02
3. Divide grams water by the molar mass of water:
.77g (1mol/18.02g) ~ .0427 mols
4. Use the mole ratios to find the number of hydrogen peroxide moles:
2H^2O^2---O^2+ 2H^2O; they are the same.
5. Multiply moles by molar mass of hydrogen peroxide.
.0427 mol (34.02g/1mol)
6. Answer
1.454g H^2O^2

Just follow the steps in this worked example.

http://www.jiskha.com/science/chemistry/stoichiometry.html

thanks bae

To solve this problem, we can use the concept of stoichiometry. Stoichiometry allows us to relate the quantities of reactants and products in a chemical reaction.

The balanced chemical equation for the decomposition of hydrogen peroxide is:

2 H2O2 → 2 H2O + O2

From the balanced equation, we can see that for every 2 moles of H2O2 decomposed, we obtain 2 moles of H2O and 1 mole of O2.

Now, let's calculate the molar mass of water (H2O) to determine the number of moles produced.

H: 1.01 g/mol (hydrogen)
O: 16.00 g/mol (oxygen)

Total molar mass = 2(1.01) + 16.00 = 18.02 g/mol

Now, let's calculate the number of moles of water produced:

0.77 g ÷ 18.02 g/mol ≈ 0.043 mole

Since the ratio of H2O2 to H2O is 2:2 (or 1:1), the number of moles of H2O2 decomposed must also be 0.043 mole.

Finally, let's calculate the mass of hydrogen peroxide decomposed:

Mass of H2O2 = molar mass × moles
= 34.01 g/mol × 0.043 mole
≈ 1.46 grams

Therefore, approximately 1.46 grams of hydrogen peroxide must decompose to produce 0.77 grams of water.