Part B. Cacijaq) + Na,00Jad) Cacofs) + 2 NaCKaq) Data 1. Amount of CaCO3() prec
ID: 556819 • Letter: P
Question
Part B. Cacijaq) + Na,00Jad) Cacofs) + 2 NaCKaq) Data 1. Amount of CaCO3() prec ipitate formed relative to the amount formed in test tube no. 3 Relative Amount of CaCO,(s) a. Test tube no. 1 b. Test tube no. 2 5.0 c. Test tube no. 31.0 d. Test tube no. 4 .0 5.0 e. Test tube no. 5 0 f. Test tube no. 6 0.25 Analysis 1. Complete the following table (1 mmol - 10 mol concentration of Na,cO, in 5 mL CaCl, in 5 mL Theoretical Excess Excess of Solution in of Solution in Yield CacO, in C0 in Ca in Tube Na,CO, CaCl, 1 1.0M 1.0 M 2 1.0 M 0.5 M 3 0.5 M 0.5 M 4 0.5 M 1.0M 5 0.5 M 0.1 M 6 0.5 M 0.01 M mmol mmol mmol Does the relative amount of CaCO,(s) observed in each test tube agree with the calculated theoretical yield for Caco,(s)? 2.Explanation / Answer
Analysis
Reaction,
Na2CO3 + CaCl2 --> CaCO3 + 2NaCl
1. Complete table
Tube 1,
mols Na2CO3 = 1.0 M x 0.005 L = 0.005 mol
mols CaCl2 = 1.0 M x 0.005 L = 0.005 mol
Theoretical yield CaCO3 = 0.005 mol
Excess CO3^2- = nil
Excess Ca^2+ = nil
Tube 2,
mols Na2CO3 = 1.0 M x 0.005 L = 0.005 mol
mols CaCl2 = 0.5 M x 0.005 L = 0.0025 mol
Theoretical yield CaCO3 = 0.0025 mol
Excess CO3^2- = 0.0025 mol
Excess Ca^2+ = nil
Tube 3,
mols Na2CO3 = 0.5 M x 0.005 L = 0.0025 mol
mols CaCl2 = 0.5 M x 0.005 L = 0.0025 mol
Theoretical yield CaCO3 = 0.0025 mol
Excess CO3^2- = nil
Excess Ca^2+ = nil
Tube 4,
mols Na2CO3 = 0.5 M x 0.005 L = 0.0025 mol
mols CaCl2 = 1.0 M x 0.005 L = 0.005 mol
Theoretical yield CaCO3 = 0.0025 mol
Excess CO3^2- = nil
Excess Ca^2+ = 0.0025 mol
Tube 5,
mols Na2CO3 = 0.5 M x 0.005 L = 0.0025 mol
mols CaCl2 = 0.1 M x 0.005 L = 0.0005 mol
Theoretical yield CaCO3 = 0.0005 mol
Excess CO3^2- = 0.002 mol
Excess Ca^2+ = nil
Tube 6,
mols Na2CO3 = 0.5 M x 0.005 L = 0.0025 mol
mols CaCl2 = 0.01 M x 0.005 L = 0.00005 mol
Theoretical yield CaCO3 = 0.00005 mol
Excess CO3^2- = 0.00245 mol
Excess Ca^2+ = nil
2. No the relative amount given above for the CaCO3 formed in each test tube does not match well with all the theoretical yield in each test tubes. Not all of Ca2+ is precipitated in many case.