What is the limited and excess reactant

Question

A solution of 25.0 grams potassium chromate reacts with a solution of 33.0 grams lead (||) nitrate to produce a yellow precipitate of lead(||) chromate and a solution of potassium nitrate. What is the limited and excess reactant. also need the mass of potassium nitrate produced. Amount of excess reactant that remains after reaction is complete 

Answer 1

Hi Anyssa,

I'll get you started here. First thing we not only want but need is a chemical equation so we can better understand how the reaction looks, takes place, and to understand how much mol of each compound is being produced.

 

K₂CrO_{4 (aq)} + Pb(NO₃)_{2 (aq)} -------> 2KNO_{3 (aq)} + PbCrO_{4 (s)}

 

Now, to find the limiting and excess reactant you need to take the amount of grams of Potassium chromate and grams of Lead(II) nitrate given and convert them to moles of your aqeous solution(KNO₃); convert each by using its molar mass to moles, then the mole ratio. The amount thats less will be your limiting reagent, and the one producing more is in excess.

 

25.0 g K₂CrO₄ x (1mol K2CrO4/194.19 g K2CrO4) x (2 mol KNO3/1mol K2CrO4) = ?

33.0 g Pb(NO₃)₂ x (1mol pb(no3)2/331.2 g pb(no3)2) x (2 mol KNO3/1 mol pb(no3)2) = ?

 

Since you found which is limiting reagent, take the original starting amount of grams stated in problem of the compound that produced the limiting reagent, and go from grams to molar mass to mole ratio to grams of KNO_3.

Lastly, amount excess left over is found by taking the moles of either k2cro4 or pb(no3)3 that produced the limiting reagent, using mole ratio, and then grams of the one that produces excess. So if for example k2cro4(which im not saying it is) produced L.R, then start with moles of k2cro4 which can be found from the stochiometry above and then go from there.