When 9.59 g of a certain vanadium oxide is heated in the...

When 9.59 g of a certain vanadium oxide is heated in the presence of hydrogen, water 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. \begin{equation} \begin{array}{l}{\text { a. Determine the empirical formulas for the two }} \\ {\text { vanadium oxides. }} \\ {\text { b. Write balanced equations for the steps of the reaction. }} \\ {\text { c. Determine the mass of hydrogen needed to complete }} \\ {\text { the steps of this reaction. }}\end{array} \end{equation}

When 9.59 g of a certain vanadium oxide is heated in
the presence of hydrogen, water 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.
\begin{equation}
\begin{array}{l}{\text { a. Determine the empirical formulas for the two }} \\ {\text { vanadium oxides. }} \\ {\text { b. Write balanced equations for the steps of the reaction. }} \\ {\text { c. Determine the mass of hydrogen needed to complete }} \\ {\text { the steps of this reaction. }}\end{array}
\end{equation}

Key Concepts

Redox Reactions

Redox (reduction-oxidation) reactions involve the transfer of electrons between species, leading to changes in their oxidation states. In the context of metal oxide reductions, a reducing agent such as hydrogen donates electrons to the metal oxide, resulting in the formation of the pure metal and water. Understanding redox reactions is vital for analyzing such processes in metallurgy, energy production, and environmental systems.

Mass Conservation

The principle of mass conservation states that in a chemical reaction, the total mass of the reactants equals the total mass of the products. This foundational concept is used to track the movement and transformation of mass during a reaction, making it indispensable for solving problems related to reaction yields and for verifying the correctness of experimentally determined or calculated quantities.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It involves converting between mass, moles, and number of particles using molar masses and balanced chemical equations. This concept is essential for predicting the amounts of chemicals consumed or produced during a reaction and is used to calculate necessary reactant volumes or masses in laboratory and industrial processes.

Empirical Formula Determination

This concept involves finding the simplest whole-number ratio of the elements in a compound from experimental data such as mass or percentage composition. It is a fundamental tool in chemistry that allows one to deduce chemical formulas based on the masses or moles of each element present, enabling further analysis of the compound's structure or reaction stoichiometry.

Balancing Chemical Equations

This key concept ensures that the law of conservation of mass is obeyed in a chemical reaction by having equal numbers of atoms for each element on both the reactant and product sides. Balancing chemical equations is critical for accurately representing chemical reactions, performing stoichiometric calculations, and understanding the underlying process of how reactants transform into products.

Transcript

00:01 In this question, we have a vanadium oxide compound that is reacting with hydrogen to form another vanadium oxide compound and water.

00:38 This second vanadium oxide compound is further reacted with hydrogen to produce metallic vanadium, so pure vanadium, and water.

01:08 To answer question a, we need to determine the empirical formula of these two vanadium oxide, this one and this one.

01:17 The information we have is that you start with 9 .59 grams of the first oxide.

01:28 You obtain 8 .76 grams of the second oxide.

01:35 These 8 .76 grams of the second oxide are further reacted, and we obtain 5 .38 grams of the second oxide.

01:49 Of the metallic vanadium.

01:53 So to determine empirical formula, let's start with the first one.

02:00 We know that we have contained here, we have 5 .38 grams of vanadium.

02:13 Therefore, the remaining of this weight, of this mass should be the oxygen.

02:24 Okay? that is 8 .76 minus 5 .38.

02:37 Okay? which means that we have of oxygen 8 .76 minus 5 .38.

02:50 We have 3 .38.

02:55 Okay.

02:56 Now we convert these two moles.

02:58 We divide each one by their respective atomic weight.

03:04 For vanadium is 50 .95, the atomic weight of vanadium.

03:10 We found that in the periodic table.

03:12 50 .94 grams per mole.

03:16 And for oxygen is 16 grams per mole.

03:21 Okay? grounds will cancel.

03:22 We're left with mold.

03:23 That's what we want.

03:24 The molar relationship between the vanadium atom and the oxygen atom.

03:30 For vanadium, we have 5 .38 divided by 50 .94.

03:35 We have 0 .1 and for oxygen we have 3 .38 divided by 16 0 .2 okay so to obtain whole numbers we divide first by the smallest of the two substrate in this case is a vanadium 0 .1 we have 1 if we divide by 0 .1 here and by 0 .1 here we want to have banalium 1 oxygen 2 and this is the empirical formula of the second oxide.

04:19 So this will correspond to what we assign as v.

04:25 Y o z.

04:27 Okay? now we do the same thing for the first oxide.

04:33 We have the y, sorry, we have, we know that we have of an adium, because the mass is conserved, we have 5 .38 grams of vanadium, also in the first compound, in the first oxide.

05:00 And of oxygen, we will have the total mass, 9 .59, minus whatever is vanadium, 5 .38.

05:14 Okay, grams.

05:18 So we have 5 .38.

05:22 And of oxygen, we have, 9 .59 minus 5 .38, which is 4 .21 grams...

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