contestada

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

Respuesta :

ardni313

Mass of SnF₂ produced : 35.634 g

Further explanation

Reaction

Sn+2HF⇒SnF₂+H₂

mol Sn (Ar 118.710 g/mol) :

[tex]\tt \dfrac{45}{118.710}=0.379[/tex]

mol SnF₂ = mol Sn = 0.379

mass SnF₂ (MW=156.69 g/mol) :

[tex]\tt 0.379\times 156.69=59.39~g[/tex]⇒theoretical

[tex]\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\60\%=\dfrac{actual}{59.39}\times 100\%\\\\actual=0.6\times 59.39=35.634~g[/tex]

Otras preguntas