75g of Cu(NO3)2 and 65g of Na2SO4. Balanced Equation : Cu (NO3)2 + Na2SO4 = CuSO4+ 2NaNO3 Calculate the mass of the copper product formed?
- Roger the MoleLv 71 day ago
Ignoring the fact that this reaction does not happen, because both products are entirely soluble:
Cu(NO3)2 + Na2SO4 → CuSO4 + 2 NaNO3
(75 g Cu(NO3)2) / (187.5558 g Cu(NO3)2/mol) = 0.39988 mol Cu(NO3)2
(65 g Na2SO4) / (142.0421 g Na2SO4/mol) = 0.45761 mol Na2SO4
0.39988 mole of Cu(NO3)2 would react completely with 0.39988 mole of Na2SO4, but there is more Na2SO4 present than that, so Na2SO4 is in excess and Cu(NO3)2 is the limiting reactant.
(0.39988 mol Cu(NO3)2) x (1 mol CuSO4 / 1 mol Cu(NO3)2) x
(159.6086 g CuSO4/mol) = 63.824 g = 64 g CuSO4
- hcbiochemLv 71 day ago
Convert masses to moles:
75 g / 187.56 g/mol = 0.400 mol Cu(NO3)2
65 g / 142.04 g/mol = 0.458 mol Na2SO4
Because these two react in a 1:1 ratio, Cu(NO3)2 is the limiting reactant and 0.400 mol of CuSO4 could possibly be formed.
Mass CuSO4 = 0.400 mol X 159.6 g/mol = 64 g CuSO4
Now, in reality, if this reaction was carried out in solution, no reaction would take place because all of the compounds are soluble in water.