Silver nitrate solution reacts with lithium chloride solution. identify the precipitate or precipitates formed.

agcl

lino3

agcl and lino3

all products are soluble

 AgCl

Explanation:  

When silver nitrate solution reacts with lithium chloride, 2 products are formed which are Silver Chloride (AgCl) and Lithium Nitrate (LiNO3).

But only one is maintained as precipitate and the other one gets solubilize in the solution.

Thus LiNO3 gets solubilize in the solution and AgCl remains as precipitate which does not get dissolve in the solution.

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

he would use the colour black