We need to go from grams of CuO to mL of H2SO4.
Atomic weight of CuO = 63.55 g Cu + 16 g O = 79.55 g CuO
(.80 g CuO) * (1 mol CuO / 79.55 g CuO) = .0100566 mol CuO (g CuO cancel)
Since the moles of CuO is a 1:1 ratio to H2SO4 (see balanced equation) we know that: mol CuO = mol H2SO4 or 0.0100566 mol CuO = 0.0100566 mol H2SO4
3.0 M of H2SO4 means that there is 3 mol / 1 L. So we can divide this by the moles to get L then mL of H2SO4
(0.0100566 mol H2SO4) * (1 L H2SO4 / 3 mol H2SO4) * (1000 mL H2SO4 / 1 L H2SO4) = 3.4 mL H2SO4 (mol H2SO4 and L H2SO4 cancel)
So 3.4 ml of H2SO4 is needed to react with 0.80 g of CuO.
The volume of sulfuric acid is 2,3 mL.
CuO + H2SO4 --> CuSO4 + H2O
CuO + H2SO4 --> CuSO4 + H2O
Cu + 2 H2SO4 ----> 2H2O+ CuSO4 + SO2Rebalanced by Graphyx
CuSO4 Cu + 2H2SO4 -> CuSO4 + SO2 + 2H2O
Simply put: Yes. 2HCN + CuSO4 yields Cu(CN)2 + H2SO4
Cu + H2SO4 = CuSO4 + H2
CuO + H2SO4 --> CuSO4 + H2O
CuO + H2SO4 --> CuSO4 + H2O
cuso4 +5h20 as h2so4 is acting as a dehydrating agent drawing the water out of the cuso4
The chemical reaction is:CuO + H2SO4 = CuSO4 + H2O
CuCO3(s) + H2SO4(aq) → CuSO4(aq) + CO2(g) + H2O(l)
Cu + 2 H2SO4 ----> 2H2O+ CuSO4 + SO2Rebalanced by Graphyx
CuSO4 Cu + 2H2SO4 -> CuSO4 + SO2 + 2H2O
Simply put: Yes. 2HCN + CuSO4 yields Cu(CN)2 + H2SO4
CuO + H2SO4 ----> CuSO4 + H2O The products are Copper(II) sulphate and water
No it isn't. It aquires properties from H2SO4 which is a strong acid, although cuso4 is a salt.
Did you mean Sulfuric acid? H2SO4 + CuSO4 -> H2SO4 + CuSO4 Since both anions are sulfates there will be no reaction as there is no change.