Answer:
Theoretical yield of vanadium = 1.6 moles
Explanation:
Moles of [tex]V_2O_5[/tex] = 1.0 moles
Moles of [tex]Ca[/tex] = 4.0 moles
According to the given reaction:-
[tex]V_2O_5_{(s)} + 5Ca_{(l)}\rightarrow 2V_{(l)} + 5CaO_{(s)}[/tex]
1 mole of [tex]V_2O_5[/tex] react with 5 moles of [tex]Ca[/tex]
Moles of Ca available = 4.0 moles
Limiting reagent is the one which is present in small amount. Thus, Ca is limiting reagent. (4.0 < 5)
The formation of the product is governed by the limiting reagent. So,
5 moles of Ca on reaction forms 2 moles of V
1 mole of Ca on reaction for 2/5 mole of V
4.0 mole of Ca on reaction for [tex]\frac{2}{5}\times 4[/tex] mole of V
Moles of V = 1.6 moles
Theoretical yield of vanadium = 1.6 moles
The theoretical yield of vanadium, in moles, that can be produced by the reaction of 1.0 mole of V₂O₅ with 4.0 moles of calcium based on the following chemical equation is 1.6 moles
Theoretical yield is the quantity of a product obtained from the complete conversion of the limiting reactant in a chemical reaction.
Given ;
According to the given reaction:-
V₂O₅ + 5Ca --> 2V + 5CaO
1 mole of V₂O₅ react with 5 moles of Ca
Moles of Ca available = 4.0 moles
Limiting reagent is the one which is present in small amount. Thus, Ca is limiting reagent. (4.0 < 5)
The formation of the product is governed by the limiting reagent.
So,
5 moles of Ca on reaction forms 2 moles of V
1 mole of Ca on reaction for 2/5 mole of V
4.0 mole of Ca on reaction for 2/5 x 4 mole of V
Moles of V = 1.6 moles
Theoretical yield of vanadium = 1.6 moles
Hence, The theoretical yield of vanadium, in moles, that can be produced by the reaction of 1.0 mole of V₂O₅ with 4.0 moles of calcium based on the following chemical equation is 1.6 moles
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