write a balanced chemical equation for the standard formation reaction of solid vanadium(v) oxide (v₂o₅).
write a balanced chemical equation for the standard formation reaction of solid vanadium(v) oxide (v₂o₅).
Answer
Explanation:
Step1: Recall formation reaction concept
A formation reaction is the reaction that forms 1 mole of a compound from its constituent elements in their standard states. Vanadium (V) is a solid and oxygen exists as $O_2$ gas in its standard - state.
Step2: Write the unbalanced equation
The unbalanced equation for the formation of $V_2O_5$ from its elements is $V(s)+O_2(g)\rightarrow V_2O_5(s)$.
Step3: Balance the equation
To balance the vanadium atoms, we need 2 moles of $V$ on the left - hand side. To balance the oxygen atoms, since there are 5 oxygen atoms in $V_2O_5$, we need $\frac{5}{2}$ moles of $O_2$ on the left - hand side. The balanced chemical equation is $2V(s)+\frac{5}{2}O_2(g)\rightarrow V_2O_5(s)$. We can also multiply through by 2 to get rid of the fraction, resulting in $4V(s) + 5O_2(g)\rightarrow2V_2O_5(s)$. But the standard formation reaction should produce 1 mole of the product, so the correct balanced equation for the standard formation reaction is $2V(s)+\frac{5}{2}O_2(g)\rightarrow V_2O_5(s)$.
Answer:
$2V(s)+\frac{5}{2}O_2(g)\rightarrow V_2O_5(s)$