If a solution contains 4.9g of tetraoxosulphate (VI) acid, calculate the amount of copper (II) oxide that will react with it [Cu = 64, O = 16, S = 32, H = 1...

If a solution contains 4.9g of tetraoxosulphate (VI) acid, calculate the amount of copper (II) oxide that will react with it

[Cu = 64, O = 16, S = 32, H = 1]

Answer Details

The balanced chemical equation for the reaction between tetraoxosulphate (VI) acid and copper (II) oxide is: H2SO4 + CuO → CuSO4 + H2O From the equation, we can see that one mole of CuO reacts with one mole of H2SO4. To find the amount of CuO that will react with 4.9g of H2SO4, we need to first find the number of moles of H2SO4 present: Molar mass of H2SO4 = 2(1) + 32 + 4(16) = 98g/mol Number of moles of H2SO4 = Mass of H2SO4/Molar mass of H2SO4 = 4.9g/98g/mol = 0.05mol Since one mole of CuO reacts with one mole of H2SO4, the number of moles of CuO needed will also be 0.05mol. The molar mass of CuO is: Molar mass of CuO = 64 + 16 = 80g/mol So, the mass of CuO needed to react with 0.05mol of H2SO4 is: Mass of CuO = Number of moles of CuO x Molar mass of CuO = 0.05mol x 80g/mol = 4g Therefore, 4g of CuO will react with 4.9g of H2SO4. The answer is 4.0g.