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For the balanced reaction: CaCl2 (aq) + Na2CO3 (aq) -> CaCO3 (s) + 2 NaCl (aq),...

For the balanced reaction: CaCl2 (aq) + Na2CO3 (aq) -> CaCO3 (s) + 2 NaCl (aq), calculate the moles of CaCO3 that form when 5.00 mL of 0.587 M CaCl2 (aq) reacts with 5.00 mL of 0.595 M Na2CO3 (aq). Enter your answer in the correct number of significant figures. Hint: This is limiting reactant problem.

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Answer #1

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

moles = molarity * volume in litres

           = 0.587M * 0.005L

           = 0.002935 moles of cacl2

and moles of na2co3 is = 0.595 * 0.005 = 0.002975 moles of na2co3

mole ratio of cacl2 and na2co3 is 1:1

i.e 1 mole cacl2 ---------------------------------> 1 mole caco3

       0.002935 mol--------------------------------->0.002935 mol

and 1 mole na2co3 ---------------------------------> 1 mole caco3

        0.002975 mol----------------------------------> 0.002975 mol

answer moles of caco3 are = 0.002935 moles

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