consider the balanced chemical equation 2H2O2 ---> 2H2) + )O2

A) given 20.0 g of H2O2 how many grams of water will be produced?

B) how many grams of H2O2 will you need to get 20.0 g of water ?

C) how many grams of H2O2 will you need to get 20.0 g of O2

D) for your answer in part C how many molecules of H2O2 is this?

See your previous post. This is a regular stoichiometric problem which makes it a little easier than the previous limiting reagent problem. Post your work if you get stuck.

THIS WAS NOT MEEEEEE

To answer these questions, we need to use stoichiometry, which is a method of using the coefficients in a balanced chemical equation to calculate the amounts of substances involved in a reaction.

The balanced chemical equation given is:
2H2O2 ---> 2H2 + O2

A) Given 20.0 g of H2O2, we need to determine how many grams of water (H2O) will be produced.

1. Calculate the molar mass of H2O2:
H2O2 = (2 * atomic mass of H) + (2 * atomic mass of O)
= (2 * 1.008 g/mol) + (2 * 16.00 g/mol)
= 34.016 g/mol

2. Determine the number of moles of H2O2:
moles of H2O2 = mass of H2O2 / molar mass of H2O2
= 20.0 g / 34.016 g/mol
≈ 0.588 mol

3. Since the stoichiometric ratio between H2O2 and H2O is 1:1:
moles of H2O = moles of H2O2
≈ 0.588 mol

4. Convert moles of H2O to grams:
mass of H2O = moles of H2O * molar mass of H2O
= 0.588 mol * 18.015 g/mol
≈ 10.57 g

Therefore, approximately 10.57 grams of water will be produced from 20.0 grams of H2O2.

B) Now, we need to determine how many grams of H2O2 will be needed to produce 20.0 grams of water.

1. Calculate the molar mass of H2O:
H2O = (2 * atomic mass of H) + atomic mass of O
= (2 * 1.008 g/mol) + 16.00 g/mol
= 18.015 g/mol

2. Determine the number of moles of H2O:
moles of H2O = mass of H2O / molar mass of H2O
= 20.0 g / 18.015 g/mol
≈ 1.110 mol

3. Since the stoichiometric ratio between H2O and H2O2 is 1:1:
moles of H2O2 = moles of H2O
≈ 1.110 mol

4. Convert moles of H2O2 to grams:
mass of H2O2 = moles of H2O2 * molar mass of H2O2
= 1.110 mol * 34.016 g/mol
≈ 37.73 g

Therefore, approximately 37.73 grams of H2O2 will be needed to produce 20.0 grams of water.

C) Next, we need to determine how many grams of H2O2 will be needed to produce 20.0 grams of O2.

1. Calculate the molar mass of O2:
O2 = 2 * atomic mass of O
= 2 * 16.00 g/mol
= 32.00 g/mol

2. Determine the number of moles of O2:
moles of O2 = mass of O2 / molar mass of O2
= 20.0 g / 32.00 g/mol
≈ 0.625 mol

3. Using the stoichiometric ratio between O2 and H2O2 (1:2):
Since 2 moles of H2O2 produce 1 mole of O2,
moles of H2O2 = 2 * moles of O2
≈ 2 * 0.625 mol
≈ 1.25 mol

4. Convert moles of H2O2 to grams:
mass of H2O2 = moles of H2O2 * molar mass of H2O2
= 1.25 mol * 34.016 g/mol
≈ 42.52 g

Therefore, approximately 42.52 grams of H2O2 will be needed to produce 20.0 grams of O2.

D) Lastly, to determine the number of molecules of H2O2 for the answer in part C:

1. Calculate the number of moles of H2O2:
moles of H2O2 = mass of H2O2 / molar mass of H2O2
= 42.52 g / 34.016 g/mol
≈ 1.25 mol

2. Use Avogadro's number to convert moles to molecules:
number of molecules = moles * Avogadro's number
= 1.25 mol * 6.022 x 10^23 molecules/mol
≈ 7.53 x 10^23 molecules

Therefore, the answer in part C corresponds to approximately 7.53 x 10^23 molecules of H2O2.