Answer:
Part A : amount of product (KCl) = 28.88 g
Part B : amount of product (KBr) = 46.13 g
Part C : amount of product (Cr₂O₃) = 17.3 g
Part D: amount of product (SrO) = 35.76 g
Explanation:
Part A:
Data Given:
Reaction :
2K(s) + Cl₂(g) --------> 2KCl
Amount of underline Reactant (K) = 15. 12g
amount of other reactant = more than enough
Explanation:
As the Potassium (K) is 15.12g and other reactant that is chlorine is more than enough so the K is limiting reagent.
So, amount of product depend on the amount of Potassium (K)
Now Look at Given Reaction:
2K(s) + Cl₂(g) --------> 2KCl
2mol 1mol 2mol
it shows that
2 mole of K give 2 mole of KCl
if we represent mole in grams
Then
Molar mass of K = 39 g/mol
Molar mass of KCl = (39 + 35.5)
Molar mass of KCl = 74.5 g/mol
So the look again to reaction in terms of grams
2K(s) + Cl₂(g) --------> 2KCl
2mole (39 g/mol) 2mole (74.5 g/mol)
78 g 149 g
Apply the Unity formula
78 g of Potassium ≅ 149 g of KCl
Then
15.12 g of Potassium ≅ how many g of Product (KCl)
By doing cross multiplication
X g of Product (KCl) = 149 g of KCl x 15.12 g of K / 78 g of K
X g of Product (KCl) = 149 g of KCl x 15.12 g of K / 78 g of K
X g of Product (KCl) = 28.88 g
So the amount of product (KCl) = 28.88 g
_________________________________________
Part B:
Data Given:
Reaction :
2K(s) + Br₂(g) --------> 2KBr
Amount of underline Reactant (K) = 15. 12g
amount of other reactant = more than enough
Explanation:
As the Potassium (K) is 15.12g and other reactant that is Bromine is more than enough so the K is limiting reagent.
So, amount of product depend on the amount of Potassium (K)
Now Look at Given Reaction:
2K(s) + Br₂(g) --------> 2KBr
2mol 1mol 2mol
it shows that
2 mole of K give 2 mole of KBr
if we represent mole in grams
Then
Molar mass of K = 39 g/mol
Molar mass of KBr = (39 + 80)
Molar mass of KBr = 119 g/mol
So, look again to reaction in terms of grams
2K(s) + Br₂(g) --------> 2KBr
2mole (39 g/mol) 2mole (119 g/mol)
78 g 238 g
Apply the Unity formula
78 g of Potassium ≅ 238 g of KBr
Then
15.12 g of Potassium ≅ how many g of Product (KBr)
By doing cross multiplication
X g of Product (KBr) = 238 g of KBr x 15.12 g of K / 78 g of K
X g of Product (KBr) = 238 g of KBr x 15.12 g of K / 78 g of K
X g of Product (KBr) = 46.13 g
So the amount of product (KBr) = 46.13 g
__________________________________________
Part C:
Data Given:
Reaction :
4Cr(s) + 3O₂(g) --------> 2Cr₂O₃
Amount of underline Reactant (Cr) = 15. 12g
amount of other reactant = more than enough
Explanation:
As the Chromium (Cr) is 15.12g and other reactant that is Oxygen is more than enough so the Cr is limiting reagent.
So, amount of product depend on the amount of Chromium (Cr)
Now Look at Given Reaction:
4Cr(s) + 3O₂(g) --------> 2Cr₂O₃
4mol 3mol 2mol
it shows that
4 mole of Cr give 2 mole of Cr₂O₃
if we represent mole in grams
Then
Molar mass of Cr = 52 g/mol
Molar mass of 2Cr₂O₃ = 2 [2 (52) + 3(16) ] = 2 (104+ 48)
Molar mass of 2Cr₂O₃ = 304 g/mol
So, look again to reaction in terms of grams
4Cr(s) + 3O₂(g) --------> 2Cr₂O₃
4 mol (52 g/mol) 2 mole (304 g/mol)
208 g 608 g
Apply the Unity formula
208 g of Chromium ≅ 608 g of Cr₂O₃
Then
15.12 g of Chromium ≅ how many g of Product (Cr₂O₃)
By doing cross multiplication
X g of Product (Cr₂O₃) = 238 g of Cr₂O₃ x 15.12 g of Cr / 208 g of Cr
X g of Product (Cr₂O₃) = 238 g of Cr₂O₃ x 15.12 g of Cr / 208 g of Cr
X g of Product (Cr₂O₃) = 17.3 g
So the amount of product (Cr₂O₃) = 17.3 g
________________________________________
Part D:
Data Given:
Reaction :
2Sr(s) + O₂(g) --------> 2SrO(s)
Amount of underline Reactant (Sr) = 15. 12g
amount of other reactant = more than enough
Explanation:
As the Strontium (Sr) is 15.12g and other reactant that is Oxygen is more than enough so the Sr is limiting reagent.
So, amount of product depend on the amount of Strontium (Sr)
Now Look at Given Reaction:
2Sr(s) + O₂(g) --------> 2SrO(s)
2mol 1mol 2mol
it shows that
2 mole of Sr give 2 mole of SrO
if we represent mole in grams
Then
Molar mass of Sr = 87.6 g/mol
Molar mass of 2SrO = 2 [87.6 + 16] = 2 (103.6)
Molar mass of 2SrO = 207.2 g/mol
So, look again to reaction in terms of grams
2Sr(s) + O₂(g) --------> 2SrO(s)
2 mol ( 87.6 g/mol) 2 mole (207.2 g/mol)
175.2 g 414.4 g
Apply the Unity formula
175.2 g of Strontium ≅ 414.4 g of SrO
Then
15.12 g of Strontium ≅ how many g of Product (SrO)
By doing cross multiplication
X g of Product (SrO) = 414.4 g of SrO x 15.12 g of Sr / 175.2 g of Sr
X g of Product (SrO) = 414.4 g of SrO x 15.12 g of Sr / 175.2 g of Sr
X g of Product (SrO) = 35.76 g
So the amount of product (SrO) = 35.76 g
