What is the mass grammes of hydrogen sulphide will be required to precipitate 15g Of Cooper sulphide from a copper (ii) tetraxosulphate (vi) solution
The proper IUPAC name is copper(II) sulfate (or sulphate).
H2S + CuSO4 ==> CuS + H2SO4
mols CuS needed = grams/molar mass = 15.0/95.5 = 0.157
Convert to mols H2S needed; i.e.,
0.157 mols CuS x (1 mol H2S/1 mol CuS) = 0.157 mols H2S needed.
Then grams H2S = mols H2S x molar mass H2S = ?
To determine the mass of hydrogen sulphide (H2S) required to precipitate 15g of copper sulphide (CuS) from a copper (II) tetraxosulphate (VI) solution, you need to calculate the stoichiometry of the reaction between H2S and CuS.
The balanced chemical equation for the reaction between H2S and CuS can be represented as follows:
CuSO4 + H2S → CuS + H2SO4
According to the stoichiometry of this reaction, 1 mole of CuS is formed from 1 mole of H2S. Therefore, we need to find the molar mass of CuS to determine the number of moles of H2S required.
The molar mass of CuS can be calculated by adding the atomic masses of copper (Cu) and sulfur (S):
Cu: 63.55 g/mol
S: 32.07 g/mol
Molar mass of CuS = Cu + S = 63.55 g/mol + 32.07 g/mol = 95.62 g/mol
Now, we can calculate the number of moles of CuS formed from the given mass:
Number of moles of CuS = Mass of CuS / Molar mass of CuS
Number of moles of CuS = 15 g / 95.62 g/mol
Once you calculate the number of moles of CuS formed, you will find that it requires a proportional amount of H2S in moles. In this case, the amount of H2S needed will be equal to the number of moles of CuS.
Therefore, the mass of H2S required to precipitate 15g of CuS from the solution is equal to the molar mass of H2S multiplied by the number of moles of CuS:
Mass of H2S = Number of moles of CuS * Molar mass of H2S
Remember, we determined the number of moles of CuS earlier. Now, assuming the molar mass of H2S is 34.08 g/mol, we can calculate the mass of H2S:
Mass of H2S = (15 g / 95.62 g/mol) * 34.08 g/mol
Therefore, the mass of hydrogen sulphide required to precipitate 15g of copper sulphide from the copper (II) tetraxosulphate (VI) solution is approximately equal to 5.34 grams.
To determine the mass of hydrogen sulphide needed to precipitate 15g of copper sulphide, we need to first calculate the stoichiometry of the reaction between hydrogen sulphide (H2S) and copper sulphide (CuS).
The balanced equation for the reaction is:
CuSO4 + H2S -> CuS + H2SO4
From the equation, we can see that one mole of copper sulphate (CuSO4) reacts with one mole of hydrogen sulphide (H2S) to produce one mole of copper sulphide (CuS).
Next, we need to calculate the molar mass of CuSO4, H2S, and CuS.
The molar mass of CuSO4:
Copper (Cu) = 63.55 g/mol
Sulfur (S) = 32.07 g/mol
Oxygen (O) = 16.00 g/mol (4 oxygen atoms in CuSO4)
Total molar mass = (63.55 + 32.07 + (16.00 x 4)) g/mol ≈ 159.61 g/mol
The molar mass of H2S:
Hydrogen (H) = 1.01 g/mol (2 hydrogen atoms in H2S)
Sulfur (S) = 32.07 g/mol
Total molar mass = (1.01 x 2 + 32.07) g/mol ≈ 34.09 g/mol
The molar mass of CuS:
Copper (Cu) = 63.55 g/mol
Sulfur (S) = 32.07 g/mol
Total molar mass = (63.55 + 32.07) g/mol ≈ 95.62 g/mol
Now, we can calculate the moles of CuS formed from the given mass of 15g:
Moles of CuS = Mass of CuS / Molar mass of CuS
Moles of CuS = 15g / 95.62 g/mol
Finally, we can use the stoichiometry of the reaction to determine the moles of H2S required:
Moles of H2S = Moles of CuS
To find the mass of H2S, we multiply the moles of H2S by its molar mass:
Mass of H2S = Moles of H2S x Molar mass of H2S
Therefore, the mass of hydrogen sulphide required can be calculated by substituting the values into the equation.