Consider the following reaction : 2CoCl3 (aq) +3 NaCO3 (aq) --- > Co2(CO3)3 (s)
ID: 780247 • Letter: C
Question
Consider the following reaction :
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 ?
Explanation / Answer
2CoCl3 (aq) +3 NaCO3 (aq) --- > Co2(CO3)3 (s) + 6 NaCl (aq)
ans(A)
moles of CoCl3 = 5.3/105.99 = 0.05 moles
so moles of NaCO3 (aq) = 3/2 * 0.05 =0.075 moles
so volume = molarity/volume = 0.075/0.2 = 0.375 = 375ml................ANS (A)
ans(b)
moles of Co2(CO3)3 formed = 1/2*0.05 = 0.025 moles..............ANS(B)
ans(c)
moles of CoCL3 = 0.4*0.05 = 0.02 moles
so moles of Na2CO3 = 3/2*0.02 = 0.03 moles
volume of Na2CO3 = 0.03/0.2 = 0.15 L = 150ml....................ASN(C)
ans (D)
at end of part C
moles of NaCl = 6/2*0.02 = 0.06 moles
moles of Co2(CO3)3 = 0.01 moles
ions present = Na+(aq) , Cl-(aq) , Co2(CO3)3(s)......................ANS(D)