Question

1. Calculate the moles of product produced in the ractions, if 4.00g of the hydrate was used in the reaction

FeCl₂ * 4H₂O_(aq) + H₂C₂O_4(aq) -> FeC₂O₄ * 2H₂O_(s) +2H₂O_(I) + 2HCl_(aq)

Ni(NO₂)₂ * 6H₂O_(aq) +H₂C₂O_4(aq) -> NiC₂O₄ * 2H₂O_(s) + 4H₂O_(I) + 2HNO_3(aq)

MnSO₄ * H₂O_(aq) + H₂C₂O_4(aq) + 2H₂O_(I) -> MnC₂O₄ * 3H₂O_(s) + 3H₂O_(I) + H₂SO_4(aq)

Answer 1

a) Mass of FeCl2 * 4H2O used = 4g

Molar mass of FeCl2 * 4H2O = 198.8 g mol^-1

Moles of FeCl2 * 4H2O =  Mass of FeCl2 * 4H2O used/ Molar mass of FeCl2 * 4H2O

= 4g/198.8 gmol^-1

   =0.02 mol

Moles of FeCl2 * 4H2O = 0.02 mol

FeCl2 * 4H2O(aq) + H2C2O4(aq) ----> FeC2O4 * 2H2O(s) +2H2O(I) + 2HCl(aq)

1 mol of FeCl2 * 4H2O(aq) produced 1 mol of FeC2O4 * 2H2O(s).

Hence, 0.02 mol  of FeCl2 * 4H2O(aq) will produce 0.02 mol of FeC2O4 * 2H2O(s).

Therefore, 0.02 mol of FeC2O4 * 2H2O(s) will be produced.

b)

Mass of Ni(NO2)2 * 6H2O used = 4g

Molar mass of Ni(NO2)2 * 6H2O = 258.8 g mol^-1

Moles of Ni(NO2)2 * 6H2O =  Mass of Ni(NO2)2 * 6H2O used/ Molar mass of Ni(NO2)2 * 6H2O

= 4g/258.8 gmol^-1

      =0.015 mol

Moles of Ni(NO2)2 * 6H2O = 0.015 mol

Ni(NO2)2 * 6H2O(aq) +H2C2O4(aq) -> NiC2O4 * 2H2O(s) + 4H2O(I) + 2HNO3(aq)

1 mol of Ni(NO2)2 * 6H2O(aq) produced 1 mol of NiC2O4 * 2H2O(s).

Hence, 0.015 mol  of Ni(NO2)2 * 6H2O(aq) will produce 0.015 mol of NiC2O4 * 2H2O(s).

Therefore, 0.015 mol of NiC2O4 * 2H2O(s) will be produced.

c)

Mass of MnSO4 * H2O used = 4g

Molar mass of MnSO4 * H2O = 169.01 g mol^-1

Moles of MnSO4 * H2O =  Mass of MnSO4 * H2O used/ Molar mass of MnSO4 * H2O

= 4g/169.01 gmol^-1

      =0.024 mol

Moles of MnSO4 * H2O = 0.024 mol

MnSO4 * H2O(aq) + H2C2O4(aq) + 2H2O(I) -> MnC2O4 * 3H2O(s) + 3H2O(I) + H2SO4(aq)

1 mol of MnSO4 * H2O(aq) produced 1 mol of MnC2O4 * 3H2O(s).

Hence, 0.024 mol  of MnSO4 * H2O(aq) will produce 0.024 mol of MnC2O4 * 3H2O(s).

Therefore, 0.024 mol of MnC2O4 * 3H2O(s) will be produced.