Respuesta :
Number of moles in the K2SO4 sample
= (16/1000)*1.04= 0.01664 mol
Number of moles in the Ba(NO3)2 sample
= (14.3/1000*0.880)= 0.01258 mol
Since the reaction is a 1:1 ratio between the two reactants, the limiting reagent is the one containing a smaller number of moles, namely Ba(NO3)2.
The molecular mass of BaSO4 is 137.3+(32.06+4*16.00)=233.4
Therefore the theoretical yield of Barium Sulphate is
233.4*0.01258=2.937 g
Actual yield = 2.60 g (given)
Therefore the percentage yield = 2.60/2.937=88.54%
Answer:
1. the limiting reagent is Barium Nitrate (Ba(NO3)2)
2. the theoretical yield is 2.94 g
3. the percentage yield is 88.5%
I apologize for the mistake previous to this update.
= (16/1000)*1.04= 0.01664 mol
Number of moles in the Ba(NO3)2 sample
= (14.3/1000*0.880)= 0.01258 mol
Since the reaction is a 1:1 ratio between the two reactants, the limiting reagent is the one containing a smaller number of moles, namely Ba(NO3)2.
The molecular mass of BaSO4 is 137.3+(32.06+4*16.00)=233.4
Therefore the theoretical yield of Barium Sulphate is
233.4*0.01258=2.937 g
Actual yield = 2.60 g (given)
Therefore the percentage yield = 2.60/2.937=88.54%
Answer:
1. the limiting reagent is Barium Nitrate (Ba(NO3)2)
2. the theoretical yield is 2.94 g
3. the percentage yield is 88.5%
I apologize for the mistake previous to this update.
An insoluble salt is created when solutions having soluble salts are reacted. The salt formed in the reaction which gets separated out is called insoluble salt.
The limiting agent is Barium Nitrate and theoretical and the percentage yields are:2.94 g and 88.5%.
This can be explained as:
K₂SO₄(aq) + Ba(NO₃)₂ (aq) → BaSO₄ (s) + 2 KNO ₃(aq)
- Number of moles in the K₂SO₄ sample
[tex]= \dfrac{16}{1000} \times 1.04= 0.01664 \;\text{mol}[/tex]
- Number of moles in the Ba(NO₃)₂ sample
[tex]= \dfrac{14.3}{1000} \times 0.880= 0.01258 \;\text{mol}[/tex]
The ratio between the two reactants is 1:1 so the limiting factor will be the one that is present in the less quantity. Hence, Ba(NO₃)₂ Barium Nitrate is the limiting factor.
The molecular mass of BaSO₄ = [tex]137.3+(32.06+4\times 16.00)=233.4[/tex]
- The theoretical yield of Barium Sulphate is
[tex]233.4 \times 0.01258=2.937 \text{g}[/tex]
Actual yield = 2.60 g
- The percentage yield [tex]= \dfrac{2.60}{2.937} = 88.54 \%[/tex]
Therefore, the limiting factor is Barium Nitrate.
To learn more about limiting factors, percentage yields and theoretical yield follow the link:
https://brainly.com/question/7786567
