Question

What is 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? V2O5(s) + 5Ca(l) → 2V(l) + 5CaO(s)

Answer 1

Answer:

Theoretical yield of vanadium = 1.6 moles

Explanation:

Moles of $V_2O_5$ = 1.0 moles

Moles of $Ca$ = 4.0 moles

According to the given reaction:-

$V_2O_5_{(s)} + 5Ca_{(l)}\rightarrow 2V_{(l)} + 5CaO_{(s)}$

1 mole of $V_2O_5$ 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 $\frac{2}{5}\times 4$ mole of V

Moles of V = 1.6 moles

Theoretical yield of vanadium = 1.6 moles