Question

Fe(NO3)3 reacts with K2CrO4 to produce Fe2(CrO4)3 according to the equation 2 Fe(NO3)3 (aq) + 3 K2CrO4 (aq) → Fe2(CrO4)3 (s) + 6 KNO3 (aq) (

a) Which reactant is limiting if 3.744 g of Fe(NO3)3 and 3.825 g of K2CrO4 are allowed to react?

(b) What mass of Fe2(CrO4)3 can be produced?

(c) What mass of the excess reactant remains when the reaction is complete?

Answer 1

The question of limiting reagent is solve by stoichiometric calculation. This is simple problem based on mole calculation. Pre knowledge of of molecular mass of compound is required to solve such problem.

Here is this solution of problem if you have any doubt then comment.

Ans Answer LAJ Solution Given equation is 2 Fe(NO3)2 + 3Kg croy re Crouz +6 KNU (6) O we know m. mass of Fe(NO3)2 = 241.86 om mo? m. mans by Kg coon = 194.18 gm molt Now check of limitting Agent @ @ check for Fe (voz) by stoichiometric calculation we know by en 2 moles of Fe(NO3)2 used by 3 moles of K Croy OR = of x 241.86 483-72 3x 194.18 58 2:54 3.825 (Given) (muss feoos) yed) 3.825 483-72 582.54 Tae = 3.17 om n a y but we have 3.744 gm Fe(NO 3) in our problem do re 003) z is in Excess amount Now het check for BK₂ crou by ein we know 3 moles of K₂ croy = 2 moles of Fe(NO3 3X 194.18 = 2x247.86 -582-54 483.72 Y 3.744 3.744 secsy 483.72 diye 3:744 x 582-54 7 483.72 St Ty = 4.50( Requered Amount of "Kqcoou we need 4.50 am af kg cool for 3.744.om my Fe 033 a stoo but we have only 3.825 gm of K₂ crou CamScanne su k acyou is limithing 4yent.

Answer B] Let the mass of Fearou) a produced sq. we Always take mass of Limitting Agent to find out mass of product ... by ein o we know 3 moles of Kg cooy = mole of Fez Crow) 3x194.18 = 1X 459.67 582. 54 4 59.67 3,895 o z -% 3.821 459.67 582.54 D = 3.825 1 459-67 $82:54 L = 3.01gm ?. - so mass of Fegerou, produced is 3. 01 Van Answerto mans of excess reactant [Fe(NO3)₂] will = be present and remains = Given - used = 3.744 - 3-17 = 0.574 gm Scannedy Thanks to CamScanner