Yes. They do not contain the same proportion of CuSO4, however. The moles of CuSO4 . 5 H2O are more massive.
29.8g H2O = 1.66 mol H2O Molar Mass CuSO4 * 5H2O = 249.6 g mol CuSO4 * 5H2O --> 5 mol H2O 249.6 g CuSO4 * 5H2O/1 mol CuSO4 * 5H2O Times * 1mol CuSO4 * 5H2O/5mol H2O Times* 1.66 mol H2O = 82.6 g CuSO4 * 5H2O
H2O CuSo4 Ca2So4
No, the percent by mass of copper in CuSO4 5H2O will be different than in CuSO4 because CuSO4 5H2O includes water molecules in addition to the copper sulfate compound itself. Therefore, the total mass of CuSO4 5H2O will be greater, resulting in a lower percent by mass of copper in CuSO4 5H2O compared to CuSO4.
The reaction between copper oxide (CuO) and sulfuric acid (H2SO4) produces copper sulfate (CuSO4) and water (H2O). The balanced chemical equation for this reaction is: CuO + H2SO4 -> CuSO4 + H2O
CuSO4 is copper (II) sulfate. The balanced equation for CuSO4 with water is CuSO4 + H2O reacts to become Cu+2 + HSO4-2 + OH-.
These are some possible ionic equation for CuSO4 plus H2O: Cu2+ + 6 H2O --> Cu(H2O)6 Cu(H2O)62+ + H2O <--> Cu(OH)(H2O)5+ + H3O+ This makes a solution of copper sulfate weakly acidic.
The balanced chemical equation for copper sulfate is: CuSO4 + H2O → CuSO4•5H2O
They both contain H2O
When heating copper(II) sulfate pentahydrate, the water molecules in the crystal structure are removed, leaving anhydrous copper(II) sulfate. The balanced chemical equation for this reaction is CuSO4·5H2O(s) → CuSO4(s) + 5H2O(g).
Copper sulfate dissolves in water, it does not react.
CuCO3(s) + H2SO4(aq) → CuSO4(aq) + CO2(g) + H2O(l)
CuCO3 + H2SO4 -> CuSO4 + H2O + CO2 copper(II) carbonate + sulphuric acid -> copper(II) sulphate water + carbon dioxide