Question

1)For the following reaction, 4.56 grams of iron(II) chloride are mixed with excess silver nitrate. The reaction yields 5.35 grams of iron(II) nitrate. iron(II) chloride (aq) + silver nitrate (aq) iron(II) nitrate (aq) + silver chloride (s) What is the theoretical yield of iron(II) nitrate ? grams What is the percent yield of iron(II) nitrate ? %

2) For the following reaction, 6.19 grams of iron are mixed with excess oxygen gas . The reaction yields 5.24 grams of iron(II) oxide . iron ( s ) + oxygen ( g ) iron(II) oxide ( s ) What is the theoretical yield of iron(II) oxide ? grams What is the percent yield for this reaction ? %

Answer 1

Q1. Balanced reaction : FeCl₂(aq) + 2 AgNO₃(aq) 2 AgCl(s) + Fe(NO₃)₂(aq)

Given : mass FeCl₂ reacted = 4.56 g

moles FeCl₂ reacted = (mass FeCl₂ reacted) / (molar mass FeCl₂)

moles FeCl₂ reacted = (4.56 g) / (126.75 g/mol)

moles FeCl₂ reacted = 0.0360 mol

moles Fe(NO₃)₂ formed = moles FeCl₂ reacted

moles Fe(NO₃)₂ formed = 0.0360 mol

theoretical yield Fe(NO₃)₂ = (moles Fe(NO₃)₂ formed) * (molar mass Fe(NO₃)₂)

theoretical yield Fe(NO₃)₂ = (0.0360 mol) * (179.85 g/mol)

theoretical yield Fe(NO₃)₂ = 6.47 g

Percent yield Fe(NO₃)₂ = (actual yield Fe(NO₃)₂ / theoretical yield Fe(NO₃)₂) * 100

Percent yield Fe(NO₃)₂ = (5.35 g / 6.47 g) * 100

Percent yield Fe(NO₃)₂ = 82.7 %

Q2. Balanced reaction : 2 Fe(s) + O₂(g) 2 FeO(s)

Given : mass Fe reacted = 6.19 g

moles Fe reacted = (mass Fe reacted) / (molar mass Fe)

moles Fe reacted = (6.19 g) / (55.845 g/mol)

moles Fe reacted = 0.111 mol

moles FeO formed = moles Fe reacted

moles FeO formed = 0.111 mol

theoretical yield FeO = (moles FeO formed) * (molar mass FeO)

theoretical yield FeO = (0.111 mol) * (71.84 g/mol)

theoretical yield FeO = 7.96 g

Percent yield FeO = (actual yield FeO / theoretical yield FeO) * 100

Percent yield FeO = (5.24 g / 7.96 g) * 100

Percent yield FeO = 65.8 %