Number of moles of oxygen gas produced from the complete catalyzed decomposition of 6.80ml sample of a 3.5% solution. Density of the 3.5% solution of H2O2 is 1.01g/ml
Let's find out what you know about this and work from there. Show your calculations to the point that you are stuck; otherwise, I may repeat much of what you already understand.
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...
Thank you for posting your work. The 6.868 g is correct. BUT that is only 3.5% H2O2; therefore,
6.868 x 0.035 = xx g H2O2
mols H2O2 then is xxg/molar mass H2O2.
mols O2 = 1/2 mols H2O2. Then I would round the number to two significant figures (since 3.5% has only two UNLESS you just left the zero off and the original problem was 3.50% H2O2).
I hope this helps.
BIG Help!!!!
Thanks....
You're welcome! I'm glad I could help. So, to continue solving the problem:
First, calculate the mass of H2O2 in the 6.80 ml sample using the density of the 3.5% solution:
mass of H2O2 = density x volume
m = 1.01 g/ml x 6.80 ml
m = 6.868 g
Next, calculate the moles of H2O2 using its molar mass:
moles of H2O2 = mass of H2O2 / molar mass of H2O2
However, since the solution is described as 3.5% H2O2, we need to adjust the mass of H2O2. Multiply the mass (6.868 g) by the decimal percent (0.035) to get the actual mass of H2O2 in the solution:
mass of H2O2 = 6.868 g x 0.035
mass of H2O2 = 0.240 g
Now, calculate the moles of H2O2 by dividing the mass by its molar mass, which is 34.0147 g/mol:
moles of H2O2 = mass of H2O2 / molar mass of H2O2
moles of H2O2 = 0.240 g / 34.0147 g/mol
moles of H2O2 ≈ 0.00706 mol
Finally, use the stoichiometric coefficients to determine the moles of oxygen gas produced:
moles of O2 = 1/2 moles of H2O2
moles of O2 = 1/2 x 0.00706 mol
moles of O2 ≈ 0.00353 mol
Therefore, the number of moles of oxygen gas produced from the complete catalyzed decomposition of the 6.80 ml sample is approximately 0.00353 mol.