Question

question provided in picture.I see an error in the question since BF3 reacts with NiCl2, but in the statement, it says that it reacts with NiF2 and the latter is the product. I will solve the question considering that 2 mol of BF3 formula units react with 4 mol of NiCl2 formula units.

Answer 1

The first thing to do is to balance the equation since the number of atoms in the reactants does not coincide with the number of atoms in the products.

So, the balanced equation will be:

`2BF_(3(s))+3NiCl_(2(aq))\rightarrow3NiF_2+2BCl_3`

So, 2 mol of BF3 reacts with 3 mol of NiCl2 to produce 3 mol of NiF2 and 2 mol of BCl3.

We have 4 mol of NiCl2 so this will be the excess reactant.

Therefore, BF3 will be the limiting reactant.

Before the reaction we will have:

2BF3

3NiCl2 + 1NiCl2(excess)

After the reaction we will have:

3NiF2

2BCl3

1NiCl2

You can produce 3 moles of nickel (II) fluoride and 2 moles of boron chloride.

The limiting reagent is BF3.

The excess reactant is NiCl2. there are 1 moles excess of this reactant.