Question

question 22 of 37
balance the following chemical equation (if necessary):
bf₃(s) + li₂so₃(s) → lif(s) + b₂(so₃)₃(s)

Explanation:

Step1: Balance B atoms

There is 1 B atom on the left and 2 B atoms on the right. So, put a coefficient of 2 in front of \(BF_3\).
\(2BF_3(s)+Li_2SO_3(s)\to LiF(s)+B_2(SO_3)_3(s)\)

Step2: Balance F atoms

After step 1, there are 6 F atoms on the left. So, put a coefficient of 6 in front of \(LiF\).
\(2BF_3(s)+Li_2SO_3(s)\to 6LiF(s)+B_2(SO_3)_3(s)\)

Step3: Balance Li atoms

After step 2, there are 6 Li atoms on the right. So, put a coefficient of 3 in front of \(Li_2SO_3\).
\(2BF_3(s)+3Li_2SO_3(s)\to 6LiF(s)+B_2(SO_3)_3(s)\)

Step4: Check SO₃ groups

On the left, we have \(3\times1 = 3\) \(SO_3\) groups (from \(3Li_2SO_3\)) and on the right, we have 3 \(SO_3\) groups (from \(B_2(SO_3)_3\)).

Answer:

\(2BF_3(s)+3Li_2SO_3(s)\to 6LiF(s)+B_2(SO_3)_3(s)\)