50.0 ml of 0.30 M Fecl2is mixed with 30.0 ml of 0.40 M Na2CO3 ,Determine which i

Question

50.0 ml of 0.30 M Fecl2is mixed with 30.0 ml of 0.40 M Na2CO3 ,Determine which ion limits the amount of precipitate formed and determine how many miles of precipitate will form,2-what is the molar its of each ion solution after precipitate has formed and settled out the mixture ,indicate which ion has0molarity? 50.0 ml of 0.30 M Fecl2is mixed with 30.0 ml of 0.40 M Na2CO3 ,Determine which ion limits the amount of precipitate formed and determine how many miles of precipitate will form,2-what is the molar its of each ion solution after precipitate has formed and settled out the mixture ,indicate which ion has0molarity? 50.0 ml of 0.30 M Fecl2is mixed with 30.0 ml of 0.40 M Na2CO3 ,Determine which ion limits the amount of precipitate formed and determine how many miles of precipitate will form,2-what is the molar its of each ion solution after precipitate has formed and settled out the mixture ,indicate which ion has0molarity?

Explanation / Answer

50.0 ml of 0.30 M Fecl2is mixed with 30.0 ml of 0.40 M Na2CO3 ,

Q1 Determine which ion limits the amount of precipitate formed and determine how many miles of precipitate will form,

Given:

Volume of FeCl2 = 50.0 mL

Molarity of FeCl2 = 0.3 M

Volume of Na2CO3 = 30.0 mL

Molarity of Na2CO3 = 0.40 M

Solution:

FeCl2 (aq) + Na2CO3 (aq)   -->2 NaCl (aq) + FeCO3 (s)

Net ionic equation :

Fe2+ (aq) + 2 Cl- (aq) + 2Na+ (aq) + Co3 2- (aq) ->2 Na+ (aq) + 2 Cl-(aq) + FeCO3 (s)

Now we cancel spectator ions from both side.

Fe2+(aq) + CO3 2- (aq) ----> FeCO3 (s)

We have to find limiting reactant. We need moles of each ion.

Mol Fe2+ = 1mol FeCl2   

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