Respuesta :
Best Answer: Your second question has the most involved answer that should cover how to do all the others as well.
For the reaction represented by the equation Cl2 + 2KBr ® 2KCl + Br2, how many grams of potassium chloride can be produced from 300. g each of chlorine and potassium bromide?
First check the equation is balanced
Cl2 + 2KBr -----> 2KCl + Br2
Then convert to moles, always work in moles
moles = mass / molar mass
molar mass Cl2 = 70.90 g/mol
moles Cl2 = 300. g / 70.90 g/mol
= 4.231 mol Cl2
molar mass KBr = 119.0 g/mol
moles KBr = 300. g / 119.0 g/mol
= 2.521 moles
Now determine the limiting and excess reagent. The limiting reagent is the reagent that will be all used up if the reaction goes to completion. There is not enough limiting reagent to use up all of the other reagent, the excess reagent.
The balanced equation tells you that
1 mol Cl2 needs 2 moles KBr to fully react
SO 4.231 moles of Cl2 needs (2 x 4.231) moles KBr
= 8.462 moles KBr
So to react all the Cl2 you would need 8.462 moles of KBr. But you only have 2.521 moles of KBr, which is not enough. So KBr is the limiting reagent. Cl2 is in excess.
The maximum amount of product you can acheive is if all the limiting reagent reacts. So the maximum amount of KCl you can get is from complete reaction of 2.521 moles of KBr
The balanced equation tells you that
2 moles KBr reacts to give 2 moles of KCl
which is a 1:1 ratio
Therefore 2.521 moles KBr produces 2.521 moles of KCl
mass = molar mass x moles
molar mass KCl = 74.6 g/mol
mass KCl = 2.521 mol x 74.6 g/mol
= 188 g KCl (3 sig figs)
For the first question
work out how many moles of O2 you produced. Then work out how many moles of KIO3 you must have started with in order to get this much O2
moles O2 = mass / molar mass
= 250. g / 32.00 g/mol
= 7.8125 moles O2 formed
The balanced equation tells you that
3 moles O2 are produced from 2 moles KIO3
So 1 mol O2 will form from 2/3 moles KIO3
Thus 7.8125 moles O2 forms from (2/3 x 7.8125) moles KIO3
= 5.21 moles KIO3 (3 sig figs)
For the 3rd question
You are told Mg(OH)2 is the excess reagent, so Fe2O3 is the limiting reagent, use same method as I used for the 2nd quuestion
The rest are similar.
Hope this helps.
For the reaction represented by the equation Cl2 + 2KBr ® 2KCl + Br2, how many grams of potassium chloride can be produced from 300. g each of chlorine and potassium bromide?
First check the equation is balanced
Cl2 + 2KBr -----> 2KCl + Br2
Then convert to moles, always work in moles
moles = mass / molar mass
molar mass Cl2 = 70.90 g/mol
moles Cl2 = 300. g / 70.90 g/mol
= 4.231 mol Cl2
molar mass KBr = 119.0 g/mol
moles KBr = 300. g / 119.0 g/mol
= 2.521 moles
Now determine the limiting and excess reagent. The limiting reagent is the reagent that will be all used up if the reaction goes to completion. There is not enough limiting reagent to use up all of the other reagent, the excess reagent.
The balanced equation tells you that
1 mol Cl2 needs 2 moles KBr to fully react
SO 4.231 moles of Cl2 needs (2 x 4.231) moles KBr
= 8.462 moles KBr
So to react all the Cl2 you would need 8.462 moles of KBr. But you only have 2.521 moles of KBr, which is not enough. So KBr is the limiting reagent. Cl2 is in excess.
The maximum amount of product you can acheive is if all the limiting reagent reacts. So the maximum amount of KCl you can get is from complete reaction of 2.521 moles of KBr
The balanced equation tells you that
2 moles KBr reacts to give 2 moles of KCl
which is a 1:1 ratio
Therefore 2.521 moles KBr produces 2.521 moles of KCl
mass = molar mass x moles
molar mass KCl = 74.6 g/mol
mass KCl = 2.521 mol x 74.6 g/mol
= 188 g KCl (3 sig figs)
For the first question
work out how many moles of O2 you produced. Then work out how many moles of KIO3 you must have started with in order to get this much O2
moles O2 = mass / molar mass
= 250. g / 32.00 g/mol
= 7.8125 moles O2 formed
The balanced equation tells you that
3 moles O2 are produced from 2 moles KIO3
So 1 mol O2 will form from 2/3 moles KIO3
Thus 7.8125 moles O2 forms from (2/3 x 7.8125) moles KIO3
= 5.21 moles KIO3 (3 sig figs)
For the 3rd question
You are told Mg(OH)2 is the excess reagent, so Fe2O3 is the limiting reagent, use same method as I used for the 2nd quuestion
The rest are similar.
Hope this helps.
