How much calcium chloride (FW=111g/mol) in grams will be needed to produce 1.00 gram of CaCO3 (FW=100g/mol) assuming Na2CO3 is in excess
Na2CO3 (aq) + CaCl2 (aq) → 2 NaCl (aq) + CaCO3 (s)
Moles of CaCO3=1.00100=0.01molesCaCO_3=\frac{1.00}{100}=0.01 molesCaCO3=1001.00=0.01moles
Moles of calcium chloride =0.01moles= 0.01moles=0.01moles
Mass of calcium chloride =0.01×111=1.11grams=0.01×111=1.11grams=0.01×111=1.11grams
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