Chapter 11, Problem 109A

When 9.59 g of a certain vanadium oxide is heated in the presence of hydrogen, and a new oxide of vanadium are formed. This new vanadium oxide has a mass of 8.76 g. When the second vanadium oxide undergoes additional heating in the presence Of hydrogen, 5.38 g of vanadium metal forms,
a. Determine the empirical formulas for the two vanadium oxides.
b. Write balanced equations for the Steps of the reaction.
c. Determine the mass of hydrogen needed to complete the steps of this reaction.

Interpretation Introduction

Interpretation : The empirical formulas of the two-vanadium oxides need to be determined.

Concept introduction : Mass of elements reacts with each other to form compounds in a definite proportion.

Answer to Problem 109A

First vanadium oxide is V₂O₅ and other is VO_2.

Explanation of Solution

Let assume first vanadium oxide,

V_wO_x and other is V_yO_z

$\begin{array}{l}VwOx+H2\to VyOz+H2O\\ 9.59g8.76g\\ VyOz+H2\to V+H2O\\ 8.765.38\end{array}$

V is 5.38 g and it is remaining constant throughout the reactions

$\begin{array}{c}MolarmassofVanadium=50.94g/mol\\ Moles=\frac{Mass}{MolarMass}\\ =\frac{5.38}{50.94}=0.1056\end{array}$

For V_wO_x

Mass of oxygen = mass of vanadium oxide − mass of vanadium

= 9.59-5.38

= 4.21

$\begin{array}{c}MolarmassofOxygen=15.99g/mol\\ Moles=\frac{Mass}{MolarMass}\\ =\frac{4.21}{15.99}=0.263\end{array}$

Vanadium and oxygen form vanadium oxide in simple ratio

Simple ration vanadium to oxygen $=\frac{0.1056}{0.263}=0.401=\frac{401}{1000}=\frac{2}{5}$

So empirical formula of this V_wO_x is V₂O_5.

For V_yO_z

Mass of oxygen = mass of vanadium oxide − mass of vanadium

= 8.76-5.38

= 3.38

$\begin{array}{c}MolarmassofOxygen=15.99g/mol\\ Moles=\frac{Mass}{MolarMass}\\ =\frac{3.38}{15.99}=0.211\end{array}$

Vanadium and oxygen form vanadium oxide in simple ratio

Simple ration vanadium to oxygen $=\frac{0.1056}{0.211}=0.500=\frac{500}{1000}=\frac{1}{2}$

So empirical formula of this V_yO_z is VO_2.

Interpretation Introduction

Interpretation:

The balanced chemical equation for each step needs to be written.

Concept introduction:

A chemical reaction is said to be balanced, if there are equal number of atoms of same elements present in the reaction.

Answer to Problem 109A

$\begin{array}{c}V2O5+H2\to 2VO2+H2O\\ VO2+2H2\to V+2H2O\end{array}$

Explanation of Solution

The balanced chemical equations will be as follows:

$\begin{array}{c}V2O5+H2\to 2VO2+H2O\\ VO2+2H2\to V+2H2O\end{array}$

From the above equations, in first reaction 1 mol of $V2O5$ reacts with 1 mol of hydrogen gas to produce 2 mol of $VO2$ and 1 mol of water.

1 mol of $VO2$ reacts with 2 mol of hydrogen gas to produce1 mol of V and 2 mol of water.

Interpretation Introduction

Interpretation:

The mass of hydrogen needed to complete both steps of reaction needs to be determined.

Concept introduction:

The relation between mass and molar mass is as follows:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Answer to Problem 109A

0.5324 g of hydrogen

Explanation of Solution

$\begin{array}{l}V2O5+H2\to 2VO2+H2O\\ 9.598.76\\ VO2+2H2\to V+2H2O\\ 8.765.38\end{array}$

$\begin{array}{c}MolarmassofV2O5=181.88g/mol\\ Moles=\frac{Mass}{MolarMass}\\ =\frac{9.59}{181.88}=0.0530\end{array}$

So, 0.0530 moles of hydrogen is require for first reaction

$\begin{array}{c}MolarmassofVO2=82.94g/mol\\ Moles=\frac{Mass}{MolarMass}\\ =\frac{8.76}{82.94}=0.1056\end{array}$

So, 2 times the 0.1056 that is 0.2112 moles of hydrogen is required for second reaction

Total moles of hydrogen = 0.0530 + 0.2112 = 0.2642

Molar mass of hydrogen = 2.0158 g/mol

Mass of hydrogen = $0.2642\times 2.015=0.5324g$