Question

question 23 of 37
balance the following chemical equation (if necessary):
li₂se(s) + so₂(g) → sse₂(g) + li₂o(s)

Explanation:

Step1: Count atoms

Count the number of each type of atom on both sides of the equation.
Left - hand side: \(Li = 2\), \(Se = 1\), \(S = 1\), \(O = 2\)
Right - hand side: \(Li = 2\), \(Se = 2\), \(S = 1\), \(O = 1\)

Step2: Balance Se atoms

To balance Se atoms, multiply \(Li_2Se\) by \(2\). The equation becomes \(2Li_2Se(s)+SO_2(g)\to SSe_2(g)+Li_2O(s)\)
Now, left - hand side: \(Li = 4\), \(Se = 2\), \(S = 1\), \(O = 2\)
Right - hand side: \(Li = 2\), \(Se = 2\), \(S = 1\), \(O = 1\)

Step3: Balance Li atoms

To balance Li atoms, multiply \(Li_2O\) by \(2\). The equation becomes \(2Li_2Se(s)+SO_2(g)\to SSe_2(g)+2Li_2O(s)\)
Now, left - hand side: \(Li = 4\), \(Se = 2\), \(S = 1\), \(O = 2\)
Right - hand side: \(Li = 4\), \(Se = 2\), \(S = 1\), \(O = 2\)

Answer:

\(2Li_2Se(s)+SO_2(g)=SSe_2(g)+2Li_2O(s)\)