Question

(A) For the following reaction, 5.83 grams of iron(II) chloride are mixed with excess silver nitrate. The reaction yields 7.31grams 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 ? __ %

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(B) According to the following reaction, how many grams of iodine are required for the complete reaction of 27.2 grams of hydrogen gas?

hydrogen (g) + iodine (s) hydrogen iodide (g)

____grams iodine

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(C) According to the following reaction, how many grams of nitrogen monoxide will be formed upon the complete reaction of 26.5 grams of oxygen gas with excess nitrogen gas?

nitrogen (g) + oxygen (g) nitrogen monoxide (g)

____grams nitrogen monoxide

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Answer 1

Answer (A)-

Given,

mass of Iron(II) Chloride = 5.83 g

Excess Silver

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

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

Actual Yield of Iron (II) Nitrate = 7.31 g

Theoretical Yield of Iron (II) Nitrate = ?

% Yield = ?

FeCl₂ (aq) + 2 AgNO₃ (Aq)$\rightarrow$ Fe(NO₃)₂ (aq) + 2 AgCl (s)

Using Stiochiometry,

It can be analyzed that for 1 mole of FeCl₂, 1 mole of Fe(NO₃)₂ is produced.

So, moles of Fe(NO₃)₂ is produced = moles of FeCl₂

We know that,

Moles = Mass/ Molar Mass

Moles of FeCl₂ in 5.83 g = 5.83 g/126.751 g/mol = 0.046 mol

So, moles of Fe(NO₃)₂ is produced = 0.046 mol

Also,

Mass = Moles *Molar Mass

Mass of Fe(NO₃)₂ is produced(Theoretical) = 0.046 mol * 179.855 g/mol

Mass of Fe(NO₃)₂ is produced(Theoretical) = 8.27 g [Answer]

Also,

% Yield = (Actual Yield/Theoretical Yield)*100

% Yield = (7.31 g/8.27g)*100

% Yield = 88.39 % [Answer]

Answer B -

Mass of Hydrogen Gas = 27.2 g

Molar mass of Hydrogen Gas = 2u

Molar mass of Iodine = 253.8089 g/mol

Mas of Iodine required = ?

H₂(g) + I₂(s)→ 2 HI(g)

Using Stiochiometry,

It can be analyzed that for 1 mole of H₂ (g), 1 mole of I₂ (g) is required.

So, Moles of I₂ (g) required = Moles of H₂ (g)

Now,

We know that,

Moles = Mass/ Molar Mass

Moles of H₂(g) = 13.6 mol

So, Moles of I₂ (g) required = 13.6 mol

Also,

Mass = Moles * Molar Mass

Mass of I₂ (g) required = 13.6 mol * 253.8089 g/mol

Mass of I₂ (g) required = 3451.80104 g [Answer]

Answer C -

Given,Mass of Oxygen Gas = 26.5 g

Nitrogen Gas Excess

Molar Mass of Oxygen Gas = 32 u

Molar mass of Nitrogen Monoxide = 30.01 g/mol

Mass of Nitrogen monoxide formed= ?

N₂ (g)+ O₂ (g) $\rightarrow$ 2 NO (g)

Using Stiochiometry,

It can be analyzed that,

For 1 mole of O₂, 2 moles of NO are produced.

So, Moles of NO produced = (1/2)* Moles of O₂

Also,

Moles = Mass/ molar Mass

Moles of O2 = 26.5 g /32 u

Moles of O2 = 0.828 mol

So, moles of NO produced = (1/2)*0.828 mol

moles of NO produced = 0.414 mol

Also,

Mass = Moles * Molar Mass

So, Mass of NO produced = 0.414 mol * 30.01 g/mol

Mass of NO produced = 12.42 g [Answer]