Question

The compound tin(II) fluoride was once a common ingredient in toothpaste. It is produced according to the following reaction: Sn(s) + 2HF(g) --> SnF2(s) + H2(g). a. What is the theoretical yield of tin (II) fluoride if 45 g of HF reacts with an excess of Sn?

Answer 1

Number of moles of HF = 45 g / 20.01 g/mol = 2.25 mole

From the balanced equation we can say that

2 mole of HF produces 1 mole of SnF2 so

2.25 mole of HF will produce

= 2.25 mole of HF *(1 mole of SnF2 / 2 mole of HF)

= 1.13 mole of SnF2

mass of 1 mole of SnF2 = 156.7068 g so

the mass of 1.13 mole of SnF2 = 177 g

Therefore, the mass of SnF2 produced would be 177 g