Question

1) For the following reaction, 50.6 grams of iron(II) chloride are allowed to react with 127 grams of silver nitrate .

iron(II) chloride(aq) + silver nitrate(aq) iron(II) nitrate(aq) + silver chloride(s)

What is the maximum amount of iron(II) nitrate that can be formed?_____________ grams

What is the FORMULA for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete? _______________grams

2) For the following reaction, 21.7 grams of iron are allowed to react with 33.9 grams of chlorine gas .

iron(s) + chlorine(g) iron(III) chloride(s)

What is the maximum mass of iron(III) chloride that can be formed?_________ grams

What is the FORMULA for the limiting reagent?

What mass of the excess reagent remains after the reaction is complete?__________ grams

please make sure to answer it clearly, as this is the third time that I had to post the same questions

Answer 1

Answer 1 -

Given,

Mass of Iron (II) Chloride (FeCl₂) = 50.6 g

Molar Mass of Iron (II) Chloride (FeCl₂) = 126.751 g/mol

Mass of Silver Nitrate (AgNO₃) = 127 g

Molar Mass of Silver Nitrate (AgNO₃) = 169.87 g/mol

Molar Mass of Iron (II) Nitrate (Fe(NO₃)₂) = 179.855 g/mol

Maximum amount of Iron (II) Nitrate Fe(NO₃)₂ = ?

Limting reagent = ?

The reaction is,

FeCl₂ (aq) + 2 AgNO₃ (aq) Fe(NO₃)₂ (aq) + 2 AgCl (s) [BALANCED]

We know that,

Moles = Mass / Molar Mass

So,

Moles of FeCl₂ = 50.6 g / 126.751 g/mol

Moles of FeCl₂ = 0.39921 mol

Moles of AgNO₃ = 127 g / 169.87 g/mol

Moles of AgNO₃ = 0.74763 mol

Now,

To Find the Limiting Reagent,

Divide the available moles by their stiochiometric coefficients, the one which is smaller is the LIMITING REAGENT.

For FeCl₂ = 0.39921 mol / 1 = 0.39921 mol

For AgNO₃ = 0.74763 mol / 2 = 0.373815 mol

So, AgNO₃ is the LIMTING REAGENT. [ANSWER]

Now,

Using Stiochiometry, it can be analyzed that, for 2 moles of AgNO₃, 1 mole of Fe(NO₃)₂ is produced. i.e.

Moles of Fe(NO₃)₂ produced = (1/2) * Moles of AgNO₃

Moles of Fe(NO₃)₂ produced = (1/2) * 0.74763 mol

Moles of Fe(NO₃)₂ produced = 0.373815 mol

Now,

Mass = Moles * Molar Mass

So,

Mass of Fe(NO₃)₂ produced = 0.373815 mol * 179.855 g/mol

Mass of Fe(NO₃)₂ produced = 67.23 g [ANSWER]

Also,

Using Stiochiometry, it can be analyzed that for 2 moles of AgNO₃, 1 mole of FeCl₂ is required. i.e.

Moles of FeCl₂ required = (1/2) * Moles of AgNO₃

Moles of FeCl₂ required = (1/2) * 0.74763 mol

Moles of FeCl₂ required = 0.373815 mol

Now,

Mass = Moles * Molar Mass

So,

Mass of FeCl₂ required = 0.373815 mol * 126.751 g/mol

Mass of FeCl₂ required = 47.38 g

Now,

Mass of Excess Reagent Left = Mass available - mass required

Mass of Excess Reagent Left = 50.6 g - 47.38 g

Mass of Excess Reagent Left = 3.22 g [ANSWER]

Answer -

Given,

Mass of Fe = 21.7 g

Mass of Cl₂ = 33.9 g

Molar Mass of Fe = 55.845 g/mol

Molar Mass of Cl₂ = 70.906 g/mol

Molar Mass of FeCl_{3 =} 162.2 g/mol

Maximum amount of FeCl₃ that can be formed = ?

Limiting Reagent = ?

Mass of Excess Reagent Left = ?

The Reaction is,

2 Fe + 3 C₂ = 2 FeCl₃

_{We know that,}

Moles = Mass / Molar mass

So,

Moles of Fe = 21.7 g / 55.845 g/mol

Moles of Fe = 0.3886 mol

Moles of Cl₂ = 33.9 g / 70.906 g/mol

Moles of Cl₂ = 0.4781 mol

To Find the LIMITNG REAGENT, divide the available moles by thir Stiochiometric coefficients, the one which is smaller is the LIMITING REAGENT.

For Fe = 0.3886 mol / 2 = 0.1943 mol

For Cl₂ = 0.4781 mol / 3 = 0.1594 mol

So, Cl₂ is the LIMITING REAGENT [ANSWER]

Now,

Using Stiochiometry, it can be analyzed that for 3 moles of Cl₂, 2 moles of FeCl₃ are produced. i.e.

Moles of FeCl₃ produced = (2/3) * Moles of Cl₂

_So,

Moles of FeCl₃ produced = (2/3) * 0.4781 mol

Moles of FeCl₃ produced = 0.3187 mol

Now,

mass = moles * Molar masss

So,

Mass of FeCl₃ produced = 0.3187 mol * 162.2 g/mol

Mass of FeCl₃ produced = 51.69 g [ANSWER]

Also,

Using Stiochiometry, it can be analyzed that for 3 moles of Cl₂, 2 moles of Fe are required. i.e.

Moles of Fe required = (2/3) * Moles of Cl₂

_So,

Moles of Ferequired = (2/3) * 0.4781 mol

Moles of Ferequired = 0.3187 mol

Now,

mass = moles * Molar masss

So,

Mass of Ferequired = 0.3187 mol * 55.845 g/mol

Mass of Ferequired = 17.797 g

Now,

Mass of Excess Reagent Left = Mass Available - Mass required

Mass of Excess Reagent Left = 21.7 g - 17.797 g

Mass of Excess Reagent Left = 3.9 g [ANSWER]