d = m/v
1.01=m/6.80 ml
m=6.868gr
coefficients of reaction: 2,2,1
two moles of H2O2 produce 1 mole of O2(g)
then 6.868gr of H202 should produce...?
I try .1009524929 and .2019049859 but they are not the right values...help...I am stuck...
See answer I posted to the original problem.
To solve this problem, we'll need to use stoichiometry to find the mass of O2 produced from a given mass of H2O2.
First, we'll start with the given mass of H2O2, which is 6.868 grams. Now we need to determine the number of moles of H2O2.
To find the number of moles, we'll use the molar mass of H2O2 which is approximately 34.02 g/mol. We can calculate the number of moles using the formula:
moles = mass / molar mass
moles = 6.868 g / 34.02 g/mol
moles ≈ 0.202 mol (rounded to three decimal places)
Now, based on the stoichiometric coefficients given (2 moles of H2O2 produce 1 mole of O2), we can determine the number of moles of O2 produced.
moles of O2 = 0.202 mol of H2O2 * (1 mol O2 / 2 mol H2O2)
moles of O2 ≈ 0.101 mol (rounded to three decimal places)
To find the mass of O2 produced, we need to multiply the number of moles by the molar mass of O2, which is approximately 32.00 g/mol.
mass of O2 = moles of O2 * molar mass of O2
mass of O2 ≈ 0.101 mol * 32.00 g/mol
mass of O2 ≈ 3.23 grams (rounded to two decimal places)
Therefore, the mass of O2 produced from 6.868 grams of H2O2 is approximately 3.23 grams.
I hope this helps! Let me know if you have any further questions.