Question

A sample of an oxide of vanadium weighing \(4.589 \mathrm{~g}\) was heated with hydrogen gas to form water and another oxide of vanadium weighing \(3.782 \mathrm{~g} .\) The second oxide was treated further with hydrogen until only \(2.573 \mathrm{~g}\) of vanadium metal remained. (a) What are the simplest formulas of the two oxides? (b) What is the total mass of water formed in the successive reactions?

Short answer

Answer: The simplest formulas of the two vanadium oxides are VO and V₂O₃, and the total mass of water formed in the successive reactions is 0.452 g.

Step-by-step solution

Find the mass of oxygen in the first oxide

Let's start by finding the mass of oxygen in the first oxide. We know that the mass of the first oxide is 4.589g and the mass of the second oxide is 3.782g. Therefore, the mass of oxygen in the first oxide is: Mass of oxygen = Mass of first oxide - Mass of second oxide Mass of oxygen = 4.589g - 3.782g Mass of oxygen = 0.807g

Find the mass of oxygen in the second oxide

Now, let's find the mass of oxygen in the second oxide. We know that the mass of the second oxide is 3.782g and that the mass of vanadium metal remaining after further treatment is 2.573g. Therefore, the mass of oxygen in the second oxide is: Mass of oxygen = Mass of second oxide - Mass of vanadium metal Mass of oxygen = 3.782g - 2.573g Mass of oxygen = 1.209g

Calculate the moles of vanadium and oxygen in each oxide

To find the simplest formula of each oxide, we need to find the mole ratio between vanadium and oxygen. First, we need to find the moles of vanadium and oxygen in each oxide using their molar masses (atomic weights): Moles of vanadium = Mass of vanadium / Molar mass of vanadium Moles of oxygen = Mass of oxygen / Molar mass of oxygen Using the atomic weights: Molar mass of vanadium = 50.94 g/mol and Molar mass of oxygen = 16.00 g/mol For the first oxide: Moles of vanadium = 2.573 g / 50.94 g/mol = 0.0505 mol Moles of oxygen = 0.807 g / 16.00 g/mol = 0.0504 mol For the second oxide: Moles of vanadium = 2.573 g / 50.94 g/mol = 0.0505 mol Moles of oxygen = 1.209 g / 16.00 g/mol = 0.0756 mol

Find the simplest formula of each oxide

Now, we can find the mole ratio between vanadium and oxygen in each oxide and thereby find the simplest formula: Mole ratio = Moles of vanadium / Moles of oxygen For the first oxide: Mole ratio = 0.0505 mol / 0.0504 mol = 1.002 The ratio is approximately 1:1, so the simplest formula of the first oxide is V₂O₂ or simply VO. For the second oxide: Mole ratio = 0.0505 mol / 0.0756 mol = 0.668 The ratio is approximately 2:3, so the simplest formula of the second oxide is V₂O₃. Part (a) answer: The simplest formulas of the two oxides are VO and V₂O₃.

Calculate the total mass of water formed

To find the mass of water formed in both reactions, we can find the mass of oxygen used in the reduction process and use the molar mass of water to find the mass of water formed: Total mass of oxygen used = Mass of oxygen in first oxide - Mass of oxygen in second oxide Total mass of oxygen used = 0.807 g - 1.209 g Total mass of oxygen used = -0.402 g Since 1 mole of water contains 1 mole of oxygen: Moles of oxygen = Moles of water Mass of water = Moles of water × Molar mass of water Mass of water = (0.402 g / 16.00 g/mol) × 18.02 g/mol Mass of water = 0.452 g Part (b) answer: The total mass of water formed in the successive reactions is 0.452 g.