Question

85 grams of BF3 and 195.0 grams of LiSO3 react according. To the chemical equation below. Use the BCA table to help you answer the questions.2BF3+3LiSO3=B2(SO3)3+6LiFWhat mass of LiF do we expect to produce

Answer 1

Step-by-step explanation:

First, let's write the chemical equation:

`2BF_3+3LiSO_3\rightarrow B_2(SO_3)_3+6LiF.`

Now, let's find what is the limiting reactant seeing which reactant will produce less product, but first, let's find what is their number of moles using its molar mass. The molar mass of BF3 is 67.8 g/mol and the molar mass of LiSO3 is 86.9 g/mol (you can calculate the molar mass of a compound using the periodic table). The conversion will look like this:

`85\text{ g BF}_3\cdot\frac{1\text{ mol BF}_3}{67.8\text{ g BF}_3}=1.25\text{ moles BF}_3.`
`195.0\text{ g LiSO}_3\cdot\frac{1\text{ mol LiSO}_3}{86.9\text{ g LiSO}_3}=2.24\text{ moles LiSO}_3.`

So, we're going to have 1.25 moles of BF3 and 2.24 moles of LiSO3 reacting.

Now, let's see how many moles of B2(SO3)3 produce each reactant.

You can see in the equation that 2 moles of BF3 reacted produces 1 mol of B2(SO3)3:

`1.25\text{ moles BF}_3\cdot\frac{1\text{ mol B}_2(SO_3)_3}{2\text{ moles BF}_3}=0.625\text{ moles B}_2(SO_3)_3.`

And 3 moles of LiSO3 reacted produces 1 mol of B2(SO3)3:

`2.24\text{ moles LiSO}_3\cdot\frac{1\text{molB}_2(SO_3)_3}{\text{3 moles LiSO}_3}=0.747\text{ moles B}_2(SO_3)_3.`

You can realize that the limiting reactant is BF3, but we need to know what is the leftover LiSO3, so let's calculate how many moles of LiSO3 are being used to produce 0.625 moles of B2(SO3)3:

`0.625\text{ moles B}_2(SO_3)_3\cdot\frac{3\text{ moles LiSO}_3}{1\text{ mol B}_2(SO_3)_3}=1.88\text{ moles LiSO}_3.`

The leftover LiSO3 is 0.36 moles.

The next step is to find the number of moles of LiF that are being produced. You can see that 2 moles of BF3 reacted produces 6 moles of LiF:

`1.25\text{ moles BF}_3\cdot\frac{6\text{ moles LiF}}{2\text{ moles BF}_3}=3.75\text{ moles LiF.}`

Answer:

So, finally, we can complete the BCA table:

Now, let's find the mass of 3.75 moles of LiF using its molar mass which is 25.9 g/mol:

`3.75\text{ moles LiF}\cdot\frac{25.9\text{ g LiF}}{1\text{ mol LiF}}=97.13\text{ g LiF.}`

We expect to produce 97.13 g of LiF.