Question

Wollastonite, CaSiO3, can be used to prepare silicon tetrafluoride, SiF4, which is used in organic synthesis, as shown below:

If you are given 7.38 g of CaSiO3 and 8.32 g of HF, what mass of the non-limiting reagent will remain

Answer 1

ANSWER:

When CaSiO₃ is treated with HF, then we will get:

CaSiO₃(s) + 6 HF(g) SiF₄(g) + CaF₂(s) + 3 H₂O(l)

Mass of CaSiO₃ = 7.38 g

Mass of HF = 8.32 g

Molecular mass of CaSiO₃ = 172.24 g/mol

Molecular mass of HF = 20.01 g/mol

Number of moles of CaSiO₃ = (7.38 g)/(172.24 g/mol)

= 0.043 mol

Number of moles of HF = (8.32 g)/(20.01 g/mol)

= 0.416 mol

From reaction, we can see that

1 mole of CaSiO₃ requires = 6 moles of HF

0.043 mol mole of CaSiO₃ requires = (6 x 0.043) moles of HF

= 0.258 moles of HF

Now,

Number of moles of non-limiting reagent (HF) will remain = (0.416 - 0.258) moles

= 0.158 moles HF

So, mass of the non-limiting reagent (HF) will remain = 0.158 mol x 20.01 g/mol

= 3.16 g

Hence, mass of the non-limiting reagent (HF) remains is 3.16 grams.