((this is not a limiting reagent problem. Rather, this is a stoichiometry proble

ID: 780318 • Letter: #

Question

((this is not a limiting reagent problem. Rather, this is a stoichiometry problem))

2CoCl3 (aq) +3 NaCO3 (aq) --- > Co2(CO3)3 (s) + 6 NaCl (aq)

Molar masses CoCl3 : 165.28 NaCo3 : 105.99 Co2(CO3)3 : 297.89

A) what valume of 0.200 M NaCO3 (aq) would be requied to react completely with 5.30 g of CoCl3?

B)How many moles of Co2(Co3)3(s) would foemd in part A?

c) what valume of 0.200 M NaCO3 (aq) would be requied to react completely with 50.0 g mL of 0.400 M CoCl3?

D) Identifay the ions the would be present at the end of part C, and state many moles of each is in the vessel ?

2CoCl3 (aq) +3 NaCO3 (aq) --- > Co2(CO3)3 (s) + 6 NaCl (aq)

2 moles of CoCl3 requires 3 moles of NaCo3

5.30 g of CoCl3 = 5.30/165.28 moles = 0.032 moles

0.032 moles of CoCl3 requires 1.5 * 0.032 moles of NaCo3 = 0.0481 mole

molarity = moles/Volume = 0.2 = 0.0481/V

v = 0.0481/2 = 0.024 litres = 24 ml

2 moles of CoCl3 requiers to form 1 moles of Co2(Co3)3(s).. SO 0.032/2 = 0.016 moles of Co2(Co3)3(s) is formed

answer is 0.016 moles of Co2(Co3)3(s)

C)Number of moles = V*Molarity = 0.4*0.05 = 0.02 moles

so it will require 1.5 *0.02 moles of NaCo3 = 0.03 moles

Volume = n/M = 0.03/0.2 = 0.15 Litres = 150 ml

0.06 moles of Na(+), 0.06 moles of Cl(-), 0.03 mole CO3(2-) , 0.02 mole Co(+3) are the ions present

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