Consider the reaction: Li 2 S(aq) + Co(NO 3 ) 2 (aq) à 2 LiNO 3 (aq) + CoS (s) W
ID: 576982 • Letter: C
Question
Consider the reaction: Li2S(aq) + Co(NO3)2 (aq) à 2 LiNO3 (aq) + CoS (s)
What is the mole-mole factor (written as a conversion factor) for each reactant to product CoS (s)?
Calculate the molar mass of each substance in the above chemical reaction.
If 4.2 g of Li2S are placed in a flask with excess Co(NO3)2, how many grams of CoS are expected? (This is theoretical yield)
The reagents in this reaction are aqueous solutions. If 20 mL of 0.25 M Co(NO3)2
are mixed with excess Li2S solution, how many grams of CoS will be formed? How many moles of LiNO3 are expected?
If 3 mL of 1.0 M Li2S is mixed with 3 mL of 0.50 M Co(NO3)2, how many grams of CoS will be produced?
After the reaction is complete, there are 50 mL of 0.5 M LiNO3 present. How much CoS (in grams) was also produced? (No information about the reactants is needed)
Explanation / Answer
Co(NO3)2 + Li2S -------------> CoS + 2LiNO3
no of moles of Co(NO3)2 = molarity * volume in L
= 0.25*0.02 = 0.005 moles
1 mole of Co(NO3)2 react with Li2S to gives 1 mole of CoS
0.005 moles of Co(NO3)2 react with Li2S to gives 0.005 moles of CoS
mass of CoS = no of moles * gram molar mass
= 0.005*91 = 0.455g
1 mole of Co(NO3)2 react with Li2S to gives 2 mole of LiNO3
0.005 moles of Co(NO3)2 react with Li2S to gives 2*0.005/1 =0.01 moles of LiNo3
no of moles of Li2S = molarity * volume in L
= 1*0.003 = 0.003 moles
no of moles of C0(NO3)2 = molarity * volume in L
= 0.5*0.003 = 0.0015 moles
1 mole of Co(NO3)2 react with 1 mole of Li2S
Co(NO3)2 is limiting reactant
1 mole of Co(NO3)2 react with Li2S to gives 1 mole of CoS
0.0015 mole of Co(NO3)2 react with Li2S to gives 0.0015 moles of CoS
mass of CoS = no of moles * gram molar mass
= 0.0015*91 = 0.1365g
no of moles of LiNO3 = molarity * volume in L
= 0.5*0.05 = 0.025 moles
LiNO3 : CoS = 2:1
= 0.025 :0.0125
no of moles of CoS = 0.0125 moles
mass of CoS = no of moles * gram molar mass
= 0.0125*91 = 1.1375g