if 15.6 grams of carbon dioxide reacts with 43.0 grams of potassium hydroxide, according to the folling reaction: co2 2koh= k2co3 h2o. Calculate the mass of potassium carbonate that will be formed(theoretical yeild) and clearly identify both the limiting reactant and the excess reactant of the reaction.

Find the number of moles of each reactant and from there, use your equation to deduce which one is limiting and which one is in excess. You will then use the limiting reactant to get the number of moles of product K2CO3 produced. From the number of moles, you should be able to get the mass of the product formed.

CO2 + 2KOH ----> K2CO3 + H2O
15.6 g of CO2
43.0 g of KOH
First you have to calculate the number of moles. You already have the mass.
so you can use the formula number of moles = mass / molecular mass ( n = m/ mr )
number of moles(CO2) = 15.6/44
= 0.355 mol
number of moles(KOH) = 43.0/56
= 0.768 mol
then you need to know which one is the limiting reagent/reactant and which one is in excess. So now that you know the number of moles of both the reactants we can use ratio method to determine which one is in excess.
CO2 : KOH
1 mol : 2 mol
0.355 mol : ?
then you cross multiply and end up with :-
? = 0.355 x 2
? = 0.71 mol of KOH we require. But if you check the number of moles we got before it is 0.786 mol. We only require 0.71 mol but we have 0.786 mol so that means KOH is in excess and obviously CO2 is the limiting reagent/reactant.
~ CO2 is the Limiting Reagent/Reactant.
~ KOH is the excess Reagent/Reactant.
then the question also also requires the mass of K2CO3
so we use the formula Mass = number of moles x molecular mass
in order to know the number of moles in K2CO3 we use the ratio method with the limiting reagent/reactant because it depends upon the limiting reagent/reactant ( I guess your chemistry teacher must have already told you that :P )
So then :-
CO2 : K2CO3
1 mol : 1 mol
0.355 mol : ?
then crossm multiply again and you end up with 0.355 moles of K2CO3.
now we have the number of moles so now use the formula Mass = number of moles x molecular mass
Mass(K2CO3) = 0.355 x 138
= 48.99 g
= 50 g ( rounded to whole number)

I think that's the right answer :) but ask your chem teacher if its totally right cause I'm just in like 10th grade so I could be wrong ( I hope not ) xD

CO2 + 2KOH ----> K2CO3 + H2O
15.6 g of CO2
43.0 g of KOH
First you have to calculate the number of moles. you already have the mass.
so you can use the formula number of moles = mass / molecular mass ( n = m/ mr )
number of moles(CO2) = 15.6/44
= 0.355 mol
number of moles(KOH) = 43.0/56
= 0.768 mol
then you need to know which one is the limiting reagent/reactant and which one is in excess. so now that you know the number of moles of both the reactants we can use ratio method to determine which one is in excess.
CO2 : KOH
1 mol : 2 mol
0.355 mol : ?
then you cross multiply and end up with :-
? = 0.355 x 2
? = 0.71 mol of KOH we require. but if you check the number of moles we got before it is 0.786 mol. we only require 0.71 mol but we have 0.786 mol so that means KOH is in excess and obviously CO2 is the limiting reagent/reactant.
~ CO2 is the Limiting Reagent/Reactant.
~ KOH is the excess Reagent/Reactant.
then the question also also requires the mass of K2CO3
so we use the formula Mass = number of moles x molecular mass
in order to know the number of moles in K2CO3 we use the ratio method with the limiting reagent/reactant because it depends upon the limiting reagent/reactant ( i guess your chemistry teacher must have already told you that :P )
So then :-
CO2 : K2CO3
1 mol : 1 mol
0.355 mol : ?
then crossm multiply again and you end up with 0.355 moles of K2CO3.
now we have the number of moles so now use the formula Mass = number of moles x molecular mass
Mass(K2CO3) = 0.355 x 138
= 48.99 g
= 50 g ( rounded to whole number)

i think thats the right answer :) but ask your chem teacher if its totally right cause im just in like 10th grade so I could be wrong ( I hope not ) xD

It's good, until the last final line. 48.99 g becomes 49 g rounded to the nearest whole number ;)

@ Unknown008
Thanks xD :)

You're welcome :)