Respuesta :
Answer:
See below explanation
Explanation:
For balanced reaction :
2 AlBr₃ + 3 K₂SO₄ → 6 KBr + Al₂(SO₄)₃
Molar Mass AlBr₃ = 267 g/mol
Molar Mass K₂SO₄ = 174 g/mol
Molar Mass KBr = 119 g/mol
If we have : 250 g AlBr₃ and 200 gr 200 g K₂SO₄ ,
534 g AlBr₃ reacts with 522 g K₂SO₄
250 g AlBr₃ = 244.38 g K₂SO₄ , so K₂SO₄ is the limiting reagent (we have less amount than required)
For theorical yield of KBr, we use the limiting reagent (K₂SO₄):
522 g K₂SO₄ should give us 714 g KBr (theorical)
200 g K₂SO₄ = 273.56 g KBr (real)
% theorical yield = (real yield/theorical yield)x100%
(273.56/714)x100 = 38.3 % for KBr
Finally, to obtain the excess reactant (AlBr₃) :
522 g K₂SO₄ reacts with 534 g AlBr₃ (all reaction is complete)
200 g K₂SO₄ = 204.60 g AlBr₃ (so all reaction would be complete)
As we have 250 g AlBr₃ : 250 g - 204.60 g = 45.4 g AlBr₃ is the excess
Answer:
A. The balanced equation is given below:
2AlBr3+ 3K2SO4 —> 6KBr + Al2(SO4)3
B. K2SO4
C. theoretical yield of KBr is 273.56g
D. 45.4g
Explanation:
The equation for the reaction. This is illustrated below:
AlBr3 + K2SO4 —> KBr + Al2(SO4)3
A. Balancing the question
AlBr3 + K2SO4 —> KBr + Al2(SO4)3
There 3 atoms of SO4 on right side and 1 atom on left side. It can be balance by putting 3 in front of K2SO4 as shown below:
AlBr3 + 3K2SO4 —> KBr + Al2(SO4)3
There are 6 atoms of K on the left side and 1 atom on the right side. It can be balance by putting 6 in front of KBr as shown below:
AlBr3 + 3K2SO4 —> 6KBr + Al2(SO4)3
There are 3 atoms of Br on the left side and 6 atoms on the right side. It can be balance by putting 2 in front of AlBr3 as shown belowb:
2AlBr3+ 3K2SO4 —> 6KBr + Al2(SO4)3
Now the equation is balanced.
B. To determine the limiting reactant, do the following:
B. Step 1:
Determination of the masses of AlBr3 and K2SO4 that reacted from the balanced equation. This is illustrated below:
2AlBr3+ 3K2SO4 —> 6KBr + Al2(SO4)3
Molar Mass of AlBr3 = 27 + (3x80) = 27 + 240 = 267g/mol
Mass of AlBr3 from the balanced equation = 2 x 267 = 534g
Molar Mass of K2SO4 = (39x2) + 32 + (16x4) = 78 + 32 + 64 = 174g/mol
Mass of k2SO4 from the balanced equation = 3 x 174 = 522g
From the balanced equation above, 534g of AlBr3 reacted 522g of K2SO4.
B. Step 2:
Determination of the limiting reactant.
2AlBr3+ 3K2SO4 —> 6KBr + Al2(SO4)3
A careful observation of the equation above, shows that K2SO4 is the limiting reactant.
Now let us check if this observation is true.
Consider using all the 200g of K2SO4 given to see if there are any leftover for AlBr3. This is illustrated below:
From the balanced equation above, 534g of AlBr3 reacted 522g of K2SO4.
Therefore, Xg of AlBr3 will react with 200g of K2SO4 i.e
Xg of AlBr3 = (534 x 200)/522
Xg of AlBr3 = 204.6g
Now we can see clearly that there are leftover for AlBr3 as only 204.6g reacted out of 250g that was given
Therefore, K2SO4 is the limiting reagent.
C. Determination of the theoretical yield of KBr.
2AlBr3+ 3K2SO4 —> 6KBr + Al2(SO4)3
Molar Mass of KBr = 39 + 80 = 119g/mol
Mass of KBr from the balanced equation = 6 x 119 = 714g
Molar Mass of K2SO4 = (39x2) + 32 + (16x4) = 78 + 32 + 64 = 174g/mol
Mass of k2SO4 from the balanced equation = 3 x 174 = 522g
The theoretical yield of KBr is obtained as follow:
From the balanced equation above,
522g of K2SO4 produced 714g of KBr.
Therefore, 200g of K2SO4 will produce = (200 x 714)/522 = 273.56g of KBr
Therefore, the theoretical yield of KBr is 273.56g
D. Determination of the leftover of the excess reactant. This is illustrated below:
The excess reactant is AlBr3.
Mass of AlBr3 given = 250g
Mass of AlBr3 that reacted = 204.6g
Mass of AlBr3 leftover =?
Mass of AlBr3 leftover = (Mass of AlBr3 given) - (Mass of AlBr3 that reacted)
Mass of AlBr3 leftover = 250 - 204.6
Mass of AlBr3 leftover = 45.4g
