an unbalanced chemical equation for the reaction of boron fluoride with lithium sulfite is shown below.\nbf₃…
an unbalanced chemical equation for the reaction of boron fluoride with lithium sulfite is shown below.\nbf₃ + li₂so₃ → b₂(so₃)₃ + lif\nwhat is the coefficient of lithium fluoride in the balanced chemical reaction?\n1\n3\n4\n6
Answer
Explanation:
Step1: Balance boron atoms
To balance the boron atoms, we need 2 moles of $BF_3$ on the left - hand side since there are 2 boron atoms in $B_2(SO_3)_3$ on the right - hand side. So the equation becomes $2BF_3+Li_2SO_3\rightarrow B_2(SO_3)_3 + LiF$.
Step2: Balance sulfate groups
There are 3 sulfate groups in $B_2(SO_3)_3$, so we need 3 moles of $Li_2SO_3$ on the left - hand side. The equation is now $2BF_3 + 3Li_2SO_3\rightarrow B_2(SO_3)_3+LiF$.
Step3: Balance lithium and fluorine atoms
On the left - hand side, we have 6 lithium atoms from $3Li_2SO_3$ and 6 fluorine atoms from $2BF_3$. So we need 6 moles of $LiF$ on the right - hand side to balance the lithium and fluorine atoms. The balanced equation is $2BF_3+3Li_2SO_3\rightarrow B_2(SO_3)_3 + 6LiF$.
Answer:
6