Answer
The balanced equation for this reaction is:
$V_2O_5(s) + 2H_2(g) -- \gt V_2O_3(s) + 2H_2O(l)$
Work Step by Step
1. Identify the chemical formula of each reactant and product:
Solid vanadium (V) oxide: $V_2O_5(s)$
Hydrogen gas: $H_2(g)$
Solid vanadium (III) oxide: $V_2O_3(s)$
Liquid water: $H_2O(l)$
2. Write the equation:
$V_2O_5(s) + H_2(g) -- \gt V_2O_3(s) + H_2O(l)$
3. Balance the equation:
$V:$ Products: 2; Reactants: 2;
$H:$ Products: 2; Reactants: 2;
$O:$ Products: 4; Reactants: 5: Unbalanced:
To balance the number of oxygens in each side, we should put a 2 as the coefficient of $H_2O$.
$V_2O_5(s) + H_2(g) -- \gt V_2O_3(s) + 2H_2O(l)$
But, now the number of hydrogens is unbalanced. To fix this problem we can put "2" as the coefficient of $H_2$:
$V_2O_5(s) + 2H_2(g) -- \gt V_2O_3(s) + 2H_2O(l)$
