Compare moles of calcium carbonate to moles of sodium carbonate based on balanced equation to calculate moles of sodium carbonate required.

Lab Report #10 Synthesis of Calcium Carbonate

Questions
Order the steps required to predict the volume (in mL) of 0.100
M sodium carbonate needed to produce 1.00 g of calcium carbonate. There is an excess of calcium chloride.
Step 1 : Convert mass of calcium carbonate to moles of calcium carbonate.
Step 2 : Compare moles of calcium carbonate to moles of sodium carbonate based on balanced equation to calculate moles of sodium carbonate required.
Step 3 : Compute the volume of sodium carbonate solution required.
Step 4 : Convert the volume of sodium carbonate solution required from liters to milliliters.
Calculate the volume (in mL) of 0.100
M needed to produce 1.00 g of
. There is an excess of .
Volume of sodium carbonate = 99.9 mL
Collected Lab Data
Collected
Volume sodium carbonate (mL) 99.0
Molarity sodium carbonate ( M ) 0.10
Volume calcium chloride (mL) 100.0
Molarity calcium chloride ( M ) 0.10
Observations
Mixture changed in appearance
Mass filter paper (g) 0.26
Mass filter paper + precipitate (g) 1.17
Calculated
Observed mass calcium carbonate (g) 0.91
Identify limiting reactant Sodium carbonate
Na2CO3
CaCO3(s)
CaCl2
9/28/22, 9:55 PM Laboratory Simulation https://newconnect.mheducation.com 2/2
Expected mass calcium carbonate (g) 0.99 Percent yield (%) 93

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