Question

How much tin II fluoride will be made when reacting 45.0 grams of tin with an excess of hydrofluoric acid if the percent yield for the reaction is 60%

Sn + HF = SnF2 + H2

Answer 1

Mass of SnF₂ produced : 35.634 g

Further explanation

Reaction

Sn+2HF⇒SnF₂+H₂

mol Sn (Ar 118.710 g/mol) :

`\tt (45)/(118.710)=0.379`

mol SnF₂ = mol Sn = 0.379

mass SnF₂ (MW=156.69 g/mol) :

`\tt 0.379* 156.69=59.39~g`⇒theoretical

`\tt \%yield=(actual)/(theoretical)* 100\%\\\\60\%=(actual)/(59.39)* 100\%\\\\actual=0.6* 59.39=35.634~g`