To determine the amount of copper(I) sulfate formed, we first need to write out the balanced chemical equation for the reaction:
Cu2O + H2SO4 -> CuSO4 + H2O
From the balanced chemical equation, we see that the molar ratio between Cu2O and CuSO4 is 1:1. This means that for every 1 mole of Cu2O reacted, 1 mole of CuSO4 is formed.
First, calculate the molar mass of Cu2O:
Cu: 63.55 g/mol
O: 16.00 g/mol
Cu2O: 2(63.55) + 16.00 = 127.10 g/mol
Now, calculate the number of moles of Cu2O:
5.37 g / 127.10 g/mol = 0.0422 mol
Since the molar ratio between Cu2O and CuSO4 is 1:1, the number of moles of CuSO4 formed is also 0.0422 mol.
Finally, calculate the mass of CuSO4 formed:
Molar mass of CuSO4: Cu = 63.55 g/mol, S = 32.07 g/mol, O = 16.00 g/mol
63.55 + 32.07 + 4(16.00) = 159.55 g/mol
Mass of CuSO4 formed: 0.0422 mol x 159.55 g/mol = 6.73 g
Therefore, 6.73 grams of copper(I) sulfate will be formed in this double displacement reaction.
When 5.37 g of the black copperIl) oxide are mixed with excess sulfuric acid, the solution turns a clear blue, indicating the formation of copper(Il) sulfate. How many grams of copper(I) sulfate will be formed in this double displacement reaction?
1 answer