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

Question

asked Aug 19, 2021 183k views

1 Answer

Hepifish · Answer 1 · 2021-08-26T01:15:24+0000

Mass of SnF₂ produced : 35.634 g

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

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