Experiment 7 1, Table 1i Synthesis of Calcium Carbonate Lab Report: Becker Moles

ID: 548511 • Letter: E

Question

Experiment 7 1, Table 1i Synthesis of Calcium Carbonate Lab Report: Becker Moles (g) of Na COp added to 100 m 10.0 milliners Moles of CaCle added to 100 mi beaker produced in the 100 ml beaker Mass (g) of CaCo, that theoretically will be will be produced in the 100 ml beaker Table 2: Filtration and Drying of Calcium Carbonate Mass (g) of fiter paper Mass (g) of watch glass Combined mass (g) of filter paper plus 6 3329 56.349 Combined Mass (g) of Watch Glass, Filter Paper, and Ca watch glass First Weighing (g) Second Weighing (g) Third Weighing (9) se.beo Fourth Weighing (g) Circle the final mass above Mass of Dry Caco, (g) Page 6 of 7

Explanation / Answer

Synthesis of Calcium Carbonate

Mass of Na2CO3 added to 100 mL beaker (g)

83.625

Mole(s) of Na2CO3 added to 100 mL beaker (mole)

(83.625 g)/(105.9888 g/mol) = 0.78899

Milliliters of 1.0 M CaCl2 added to 100 mL beaker (mL)

10.0

Mole(s) of CaCl2 added to 100 mL beaker (mole)

(10.0 mL)*(1 L/1000 mL)*(1.0 M)*(1 mol/L/1 M) = 0.01

Mole(s) of CaCO3 that theoretically will be produced in the 100 mL beaker (mole)

0.01(see explanation below)

Mass of CaCO3 that theoretically will be produced in the 100 mL beaker (g)

(0.01 mole)*(100.0869 g/mol) = 1.000869

Explanation:

Write down the balanced chemical equation.

Na2CO3 + CaCl2 --------> CaCO3 + 2 NaCl

As per the stoichiometric equation,

1 mole Na2CO3 =1 mole CaCl2 = 1 mole CaCO3.

Clearly, we have fewer mole(s) of CaCl2; hence, CaCl2 is the limiting reagent and the yield of CaCO3 is decided by the mole(s) of CaCl2.

Mass of dry CaCO3 = (combined mass of watch glass, filter paper and CaCO3 after second drying) – (combined mass of filter paper plus watch glass) = (58.295 g) – (58.299 g) = -0.004 g; clearly this result is wrong; you didn’t record the masses properly. The final mass must always be higher than the initial mass the product mass will be a positive value. Check the data again.