In: Chemistry
CuSO4·5H2O +
2NaC7H5O2 →
Cu(C7H5O2)2·2H2O
+ Na2SO4 +
3H2O |
a) Determine the number of moles of CuSO4·5H2O contained in 2.00 g of CuSO4·5H2O.
b)Calculate the mass of NaC7H5O2 you will need to react "completely" with the 2.00 g of CuSO4·5H2O.
Ans a)
Molar mass of CuSO4·5H2O is 249.69 g/mol. The number of moles of CuSO4·5H2O in 2.0g of CuSO4·5H2O can be calculated as follows:
Number of moles of CuSO4·5H2O = Weight of CuSO4·5H2O / molar mass of CuSO4·5H2O
Number of moles of CuSO4·5H2O = (2.00 g) / (249.69 g/mol)
Number of moles of CuSO4·5H2O = 0.00801 mol
Ans b)
The balanced equation of reaction of CuSO4·5H2O and NaC7H5O2 can be written as follows:
CuSO4·5H2O + 2NaC7H5O2 → Cu(C7H5O2)2·2H2O + Na2SO4 + 3H2O
Thus, one mole of CuSO4·5H2O reacts with two moles of NaC7H5O2.
Therefore, 2.00g of CuSO4·5H2O (0.00801 mol) will react with (0.00801 x 2) mol of NaC7H5O2. Thus, mass of (0.00801 x 2) mol of NaC7H5O2 can be calculated as follows:
Mass of (0.00801 x 2) mol of NaC7H5O2 = (0.00801 x 2) mol x molar mass of NaC7H5O2
Mass of (0.00801 x 2) mol of NaC7H5O2 = (0.00801 x 2) mol x 144.10 g/mol
Mass of (0.00801 x 2) mol of NaC7H5O2 = 2.31 g
Thus, the mass of NaC7H5O2 needed to react "completely" with the 2.00 g of CuSO4·5H2O is 2.31 g.