05 978 107169 1.191a 0.915 0.99D 1.081 0.1650 0001130 Male s 1. Mass of beaker (

Question

05 978 107169 1.191a 0.915 0.99D 1.081 0.1650 0001130 Male s 1. Mass of beaker (g) 2. Mass of beaker and salt mixture (g) 3. Mass of mixture (g) 4. Mass of filter paper (g) 5. Mass of filter paper and air-dried CaC204 H2O 6. Mass of air-dried CaC204 H20 7. Moles of CaC204 H20 at Go 8. 9. 10. 11. 12. 13. 14, 15, 16. 17. 18. Limiting reactant in salt mixture (write formula) Excess reactant in salt mixture (write formula) Moles of limiting reactant Mass of limiting reactant (g) Mass of excess reactant (g) Moles of excess reactant Mass % of limiting reactant Mass % of excess reactant in mixture Moles of excess reactant that reacted Mass of excess reactant that reacted (g) Mass of excess reactant that is unreacted (g) 56

Explanation / Answer

chemical equation for the reaction (without water of hydration),

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

1 mole of CaCl2 reacts with 1 mole of K2C2O4 to form 1 mole of CaC2O4

(10) moles of CaC2O4 = 0.00113 mols

moles of limiting reactant = 0.00113 mols

(11) mass of limiting reactant (CaCl2.2H2O) = 0.00113 mol x 147.014 g/mol = 0.17 g

(12) Mass of excess reactant K2C2O4.H2O = 1.191 - 0.17 = 1.021 g

(13) moles excess reactant = 1.021 g/184.231 g/mol = 0.0055 mols

(14) mass% limiting reactant = 0.17 x 100/1.191 = 14.27%

(15) mass% of excess reactant in mixture = 1.021 x 100/1.191 = 85.73%

(16) moles of excess reactant that reacted = 0.00113 mols

(17) mass of excess reactant that reacted = 0.00113 mol x 184.231 g/mol = 0.21 g

(18) mass of excess reactant that is unreacted = 1.021 - 0.21 = 0.811 g

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