What Mass In Grammes Of Hydrogen sulphide will Be Required To Precipitate 15g of Copper Sulphide from A Copper(II)tetraoxosulphate(VI)solution?
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first step: write the balanced reaction equation.
Then you can see how many moles of stuff are needed.
Just a note that copper sulfate is an accepted name for CuSO4 by IUPAC
What Mass In Grammes Of Hydrogen sulphide will Be Required To Precipitate 15g of Copper Sulphide from A Copper(II)tetraoxosulphate(VI)solution?
To determine the mass of hydrogen sulphide required to precipitate copper sulphide from a copper(II) tetraoxosulphate(VI) solution, we need to know the balanced equation for the reaction between copper(II) ions and hydrogen sulphide.
The balanced equation for the reaction is as follows:
CuSO4 + H2S → CuS + H2SO4
By examining the equation, we can see that one mole of hydrogen sulphide (H2S) reacts with one mole of copper sulphate (CuSO4) to produce one mole of copper sulphide (CuS) and one mole of sulphuric acid (H2SO4).
Now, let's calculate the molar mass of CuS (copper sulphide) and H2S (hydrogen sulphide):
- Molar mass of copper sulphide (CuS):
Cu (63.55 g/mol) + S (32.07 g/mol) = 95.62 g/mol
- Molar mass of hydrogen sulphide (H2S):
H (1.01 g/mol) + S (32.07 g/mol) = 33.08 g/mol
To find out how many moles of CuS are required to precipitate 15g of CuS, divide the given mass by the molar mass:
Moles of CuS = Mass of CuS / Molar mass of CuS
Moles of CuS = 15g / 95.62 g/mol
Now, since the reaction is 1:1 between H2S and CuS, the same number of moles of H2S will be required:
Moles of H2S = Moles of CuS
Finally, we can calculate the mass of H2S required:
Mass of H2S = Moles of H2S * Molar mass of H2S
Substitute the moles of H2S (which is equal to the moles of CuS) into the formula:
Mass of H2S = (15g / 95.62 g/mol) * (33.08 g/mol)
By performing the calculations, we will get the mass of hydrogen sulphide required to precipitate 15g of copper sulphide from the given solution.