Question

a and b

Question: What is the percentage yield of calcium carbonate formed when calcium chloride reacts with sodium carbonate? 

PART A: 

Use the balanced chemical equation below to determine: 

(a) the limiting reactant 

(b) theoretical yield (mass) of calcium carbonate 

\(\underset{\mathrm{m}=3.75 \mathrm{~g}}{\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{qq})}+\underset{\mathrm{m}=4.82 \mathrm{~g}}{\mathrm{CaCl}_{2}(\mathrm{oq})} \rightarrow \underset{\mathrm{m}=?}{\mathrm{CaCO}_{3}(\mathrm{~s})}+2 \mathrm{NaCl}_{(\mathrm{oq})}\)

Answer 1

Na2CO3(aq) + CaCl2(aq) ------------- CaCO3(s) + 2 NaCl(aq)

1 mole             1 mole                        1 mole

a)

mass of Na2CO3 = 3.75 grams

molar mass of Na2CO3 = 106.0 gram/mole

number of moles ofNa2CO3 = mass/molar mass = 3.75 / 106.0 = 0.0354 moles

mass of Cacl2 = 4.82 grams

molar mass of CaCl2 = 110.98 gram/mole

number of moles of CaCl2 = 4.82 / 110.98 = 0.0434 moles

according to equation

1 mole of Na2CO3 = 1 mole of CaCl2

number of moles of Na2CO3 is less than that of CaCl2. SO Na2CO3 is completed first during the reaction.

Hence Na2CO3 is limiting reagent.

b) Na2CO3 is limiting reagent

according ot equation 1 mole of Na2CO3 = 1 mole of CaCO3

0.0354 moles of Na2CO3 = ?

                                                 = 0.0354 x 1 /1 = 0.0354 moles of CacO3

number of moles of CaCO3 formed = 0.0354 moles

molar mass of CaCO3 = 100.0 gram/mole

mass of 0.0354 moles = 0.0354 x 100 = 3.54 grams

mass of CaCO3 formed = 3.54 grams

Theoritical yield of CaCO3 = 3.54 grams.