Question

effect of limiting concentration of a reactant lab answer

i need help doing the calculations for the last page. I dont understand the questions

Answer 1

Reaction: Na2CO3(aq) + CaCl2(aq) ----> CaCO3(s) + 2NaCl(aq)

Tube No. 1:

Millimoles of Na2CO3 = 5 mL * 1 mmol/1 mL = 5 mmol

Millimoles of CaCl2 = 5 mL * 1 mmol/1 mL = 5 mmol

Therefore, Anything (CaCl2 or Na2CO3) is the limiting reactant, i.e. there is no excess reactant.

From the reaction stoichiometry, millimoles of product possible = 5 mmol

Tube No. 2:

Millimoles of Na2CO3 = 5 mL * 1 mmol/1 mL = 5 mmol

Millimoles of CaCl2 = 5 mL * 0.5 mmol/1 mL = 2.5 mmol

Therefore, CaCl2 is considered as the limiting reactant, and Na2CO3 is the excess reactant.

From the reaction stoichiometry, millimoles of product possible = 2.5 mmol

Tube No. 3:

Millimoles of Na2CO3 = 5 mL * 0.5 mmol/1 mL = 2.5 mmol

Millimoles of CaCl2 = 5 mL * 0.5 mmol/1 mL = 2.5 mmol

Therefore, Anything (CaCl2 or Na2CO3) is the limiting reactant, i.e. there is no excess reactant.

From the reaction stoichiometry, millimoles of product possible = 2.5 mmol

Tube No. 4:

Millimoles of Na2CO3 = 5 mL * 0.5 mmol/1 mL = 2.5 mmol

Millimoles of CaCl2 = 5 mL * 1 mmol/1 mL = 5 mmol

Therefore, Na2CO3 is considered as the limiting reactant, and CaCl2 is the excess reactant.

From the reaction stoichiometry, millimoles of product possible = 2.5 mmol

Tube No. 5:

Millimoles of Na2CO3 = 5 mL * 0.5 mmol/1 mL = 2.5 mmol

Millimoles of CaCl2 = 5 mL * 0.1 mmol/1 mL = 0.5 mmol

Therefore, CaCl2 is considered as the limiting reactant, and Na2CO3 is the excess reactant.

From the reaction stoichiometry, millimoles of product possible = 0.5 mmol

Tube No. 6:

Millimoles of Na2CO3 = 5 mL * 0.5 mmol/1 mL = 2.5 mmol

Millimoles of CaCl2 = 5 mL * 0.05 mmol/1 mL = 0.25 mmol

Therefore, CaCl2 is considered as the limiting reactant, and Na2CO3 is the excess reactant.

From the reaction stoichiometry, millimoles of product possible = 0.25 mmol