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?
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.