[OrdinaryDay]

Iron(II) Chloride reacted with 148 grams of silver nitrate. Calculate the amount of grams for the products at 57 degrees Celsisus and 740 mmHg.

^ I don't quite understand how to approach this problem other than the fact that Fe(NO₃)₂ results in 63 grams and AgNO₃ results in 29 grams when temperature and pressure are excluded.

[Borek]

Start with the reaction equation.

[OrdinaryDay]

The balanced equation:

FeCl₂(aq) +2 AgNO₃---->Fe(NO₃)₂(aq)+2AgCl₂

The amount of moles:

44.4 g FeCl₂(1 mol FeCl₂/126.8 FeCl₂) =.3502 mol FeCl₂

148 g AgNO₃(1 mol/ 169.9 g)= .871 mol AgNO₃

The amount of moles needed to use up all of the .3502 mols of FeCl₂

.3502 mol FeCl₂ (2 mols AgNO₃/1 mol FeCl₂)

The rest is very obvious however the extra conditions with temp and pressure confused me..

[Borek]

AgCl not AgCl₂, but I suppose it is just a typo.

Otherwise I have no idea what you are doing - where is 44.4 g FeCl₂ from, it wasn't listed in the original question.

Extra conditions don't matter, they are here just to confuse you.