Question

DATA SHEET:
1- Mass of beaker = 66.7521g
2-Mass of beaker and salt mixture = 67.798g
3-Mass of salt mixture = 1.0465g
4-Mass of filter paper = 1.2110g
5-Mass of filter paper and CaC2O4*H2O = 1.5948g
6-Mass of Air-dried CaC2O4*H2O = 0.3838g

1- Limiting reactant in salt mixture = K2C2O4*H2O
2- Excess reactant in salt mixture = CaCl*2H2O

How do i find these questions out?


1-Moles of CaC2O4*H2O precipitated:______________________
2-Moles of limiting reactant in salt mixture(mol) = ?

- Formula of limiting hydrate_____________________

3-Moles of limiting reactant in salt mixture(g) = ?

- Formula of limiting hydrate_____________________

4-Mass of excess reactant in salt mixture (g)

- Formula of Excess hydrate_____________________

5-Percent limiting reactant in salt mixture(%) = ?

- Formula of limiting hydrate ____________________

6-Percent excess reactant in salt mixture(%) = ?

- Formula of excess hyrdate_____________________

7- Mass of excess reactant that reacted(g) =

- Formula of excess reacted_________________________

8- Mass of excess reacted , unreacted (g) ___________________________________

Answer 1

DATA SHEET:
1- Mass of beaker = 66.7521g
2-Mass of beaker and salt mixture = 67.798g
3-Mass of salt mixture = 1.0465g
4-Mass of filter paper = 1.2110g
5-Mass of filter paper and CaC2O4*H2O = 1.5948g
6-Mass of Air-dried CaC2O4*H2O = 0.3838g

1- Limiting reactant in salt mixture = K2C2O4*H2O
2- Excess reactant in salt mixture = CaCl*2H2O

How do i find these questions out?

The reaction will be

K2C2O4.H2O + CaCl2.2H2O --> CaC2O4.H2O + 2KCl + 2H2O

1-Moles of CaC2O4*H2O precipitated: Mass / molecular weight = 0.3838 / 146 = 0.00262 moles
2-Moles of limiting reactant in salt mixture(mol) = ?

Moles of limiting reactant in salt mixture(mol) = K2C2O4*H2O

Mass of Air-dried CaC2O4*H2O = 0.3838g

1mol CaCl2 reacts with 1mol K2C2O4 to produce 1mol CaC2O4

Moles of   CaC2O4 = 0.00262 moles

so moles of K2C2O4 = 0.00262

- Formula of limiting hydrate = KC2O4.H2O

3-Mass of limiting reactant in salt mixture(g) = moles X mol wt = 0.00262 X 166.2 = 0.4354 g

4-Mass of excess reactant in salt mixture (g) = total mass - mass of limiting reagent = 1.0465 - 0.4354 = 0.6111 g

4- Formula of Excess hydrate = CaCl.2H2O