Hydorgen peroxide may be prepared by the following reactions:
2NH4HSO4 ---> H2+(NH4)2S2O8
(NH4)2S2O8+2H2O ----> 2NH4HSo4+H2O2
What mass of ammonium hydrogen sulfate, Nh4HSo4 is initially required to prepare 2.00 mole of H2O2?
What mass of H2O is required?
I have checked and the reaction are ballanced. But the wording is confusing me with what I have in my notes and I'm not sure what to do next. Please advise.
You want me to work the parts you've already worked. Show what you've done and we can tell you what to do next.
This link will help too.
To find the mass of ammonium hydrogen sulfate (NH4HSO4) required to prepare 2.00 moles of H2O2, we need to make use of the balanced chemical equation:
2NH4HSO4 → H2 + (NH4)2S2O8
(NH4)2S2O8 + 2H2O → 2NH4HSO4 + H2O2
The molar ratio between NH4HSO4 and H2O2 is 2:1 based on the balanced equation.
First, let's calculate the number of moles of NH4HSO4 required to produce 2.00 moles of H2O2:
Number of moles of NH4HSO4 = 2.00 moles H2O2 ÷ 1 (from the balanced equation) = 2.00 moles NH4HSO4
Next, we'll determine the molar mass of NH4HSO4.
The molar mass of NH4HSO4 is calculated as:
(1 x Nitrogen(N)) + (4 x Hydrogen(H)) + (1 x Sulfur(S)) + (4 x Oxygen(O)) + (4 x Hydrogen(H)).
Molar mass of NH4HSO4 = (1 x 14.01 g/mol) + (4 x 1.01 g/mol) + (1 x 32.07 g/mol) + (4 x 16.00 g/mol) + (4 x 1.01 g/mol) = 132.14 g/mol.
Finally, we can calculate the mass of NH4HSO4 required:
Mass of NH4HSO4 = Number of moles of NH4HSO4 x Molar mass of NH4HSO4
Mass of NH4HSO4 = 2.00 moles x 132.14 g/mol = 264.28 g
Therefore, 264.28 grams of ammonium hydrogen sulfate (NH4HSO4) is initially required to prepare 2.00 moles of H2O2.
To find the mass of H2O required, we'll again use the balanced equation:
(NH4)2S2O8 + 2H2O → 2NH4HSO4 + H2O2
According to the balanced equation, the molar ratio of H2O to H2O2 is 2:1.
Thus, the number of moles of H2O required will be half the number of moles of H2O2:
Number of moles of H2O = 2.00 moles of H2O2 ÷ 2 (from the balanced equation) = 1.00 mole of H2O
The molar mass of H2O is 18.02 g/mol.
Mass of H2O = Number of moles of H2O x Molar mass of H2O
Mass of H2O = 1.00 mole x 18.02 g/mol = 18.02 g
Therefore, 18.02 grams of H2O is required.
To find the mass of ammonium hydrogen sulfate (NH4HSO4) required to prepare 2.00 moles of H2O2, we need to use stoichiometry.
Let's start with the balanced chemical equation:
(NH4)2S2O8 + 2H2O -> 2NH4HSO4 + H2O2
From the equation, we can see that the ratio between (NH4)2S2O8 and NH4HSO4 is 1:2. This means that for every 1 mole of (NH4)2S2O8, we get 2 moles of NH4HSO4.
Therefore, we need to convert the moles of H2O2 into moles of NH4HSO4 using the stoichiometry of the reaction.
Given that we have 2.00 moles of H2O2, we can set up the following conversion:
2.00 moles H2O2 * (2 moles NH4HSO4 / 1 mole H2O2) = 4.00 moles NH4HSO4
Now that we have the number of moles of NH4HSO4 required, we can convert it to mass.
To do that, we need to know the molar mass of NH4HSO4, which is:
NH4 = 18.04 g/mol, H = 1.01 g/mol, S = 32.07 g/mol, and O = 16.00 g/mol.
NH4HSO4 = 18.04 + 1.01 + 32.07 + (4 * 16.00) = 132.14 g/mol
Now, we can calculate the mass of NH4HSO4 required:
4.00 moles NH4HSO4 * 132.14 g/mol = 528.56 grams of NH4HSO4
Therefore, to prepare 2.00 moles of H2O2, you would need 528.56 grams of ammonium hydrogen sulfate (NH4HSO4).
To find the mass of H2O required, we need to use the stoichiometry of the reaction:
From the balanced equation, you can see that for every 1 mole of (NH4)2S2O8, we get 1 mole of H2O.
Therefore, the moles of H2O required is equal to the moles of (NH4)2S2O8.
Since the reaction is not starting with (NH4)2S2O8, we cannot determine its moles or mass from the given information.