Question

The tungsten metal used for filaments in light bulbs is made by reaction of tungsten(VI) oxide with hydrogen: \(\mathrm{WO}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{W}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)\) The above reaction was performed and produced \(5.00 \mathrm{~g}\) of tungsten. (a) How many moles of tungsten were formed? (b) How many moles of tungsten(VI) oxide and hydrogen were required to produce the \(5.00 \mathrm{~g}\) of tungsten? (c) How many grams of tungsten(VI) oxide, and how many grams of hydrogen must you start with to prepare \(5.00 \mathrm{~g}\) of tungsten? (For \(\mathrm{WO}_{3},\) molecular mass \(=231.8\) amu. \()\)

Short answer

5.00 g tungsten corresponds to 0.0272 moles W, requiring 6.31 g WO₃ and 0.165 g H₂.

Step-by-step solution

Calculate Moles of Tungsten (W) Formed

To find out how many moles of tungsten were formed, we need to use tungsten's molar mass. The atomic mass of tungsten (W) is approximately 183.8 g/mol. Given that 5.00 g of tungsten was produced, we use the formula: \[ \text{moles of tungsten} = \frac{\text{mass}}{\text{molar mass}} = \frac{5.00 \text{ g}}{183.8 \text{ g/mol}} \approx 0.0272 \text{ moles} \]

Determine Moles of WO3 and H2 Required

According to the balanced chemical equation: \( \mathrm{WO}_{3}(s) + 3 \mathrm{H}_{2}(g) \rightarrow \mathrm{W}(s) + 3 \mathrm{H}_{2} \mathrm{O}(g) \)One mole of WO₃ reacts with 3 moles of H₂ to produce one mole of W. Since 0.0272 moles of tungsten were produced, the same amount of WO₃ and 3 times that amount of H₂ were needed.\[ \text{moles of WO}_3 = 0.0272 \text{ moles} \]\[ \text{moles of H}_2 = 3 \times 0.0272 = 0.0816 \text{ moles} \]

Calculate Mass of WO3 Required

To find the mass of WO₃ required, we use the number of moles calculated and the molar mass of WO₃, which is 231.8 g/mol.\[ \text{mass of WO}_3 = \text{moles of WO}_3 \times \text{molar mass of WO}_3 \]\[ \text{mass of WO}_3 = 0.0272 \times 231.8 \approx 6.31 \text{ g} \]

Calculate Mass of H2 Required

To determine the mass of H₂ needed, use the moles of H₂ calculated and its molar mass (approximately 2.02 g/mol).\[ \text{mass of H}_2 = \text{moles of H}_2 \times \text{molar mass of H}_2 \]\[ \text{mass of H}_2 = 0.0816 \times 2.02 \approx 0.165 \text{ g} \]

Key concepts

Chemical Reactions

A chemical reaction involves the transformation of substances through the breaking and forming of chemical bonds. In the case of tungsten metal production, the reaction involves tungsten(VI) oxide (\( \mathrm{WO}_3 \)) and hydrogen (\( \mathrm{H}_2 \)). During the reaction, these substances undergo a transformation to yield pure tungsten metal (\( \mathrm{W} \)) and water vapor (\( \mathrm{H}_2 \mathrm{O} \)).

• Reactants: These are the starting materials in a chemical reaction. Here, they are tungsten(VI) oxide and hydrogen.
• Products: These are the substances produced as a result of the reaction. In this instance, they are tungsten metal and water vapor.
• Conservation of Mass: In any chemical reaction, the total mass of the reactants equals the total mass of the products due to the conservation of mass principle.

Understanding the specifics of how reactants convert to products helps in analyzing and predicting the results of chemical reactions.

Molar Mass Calculations

Calculating molar mass is a crucial step in stoichiometry that helps convert between mass and moles. Molar mass is the sum of atomic masses of all atoms in a compound.

• Atomic Mass: Every element has an atomic mass, typically found on the periodic table. For example, the atomic mass of tungsten (\( \mathrm{W} \)) is approximately 183.8 g/mol.
• Molar Mass of a Compound: For compounds, such as tungsten(VI) oxide, the molar mass is calculated by adding the masses of constituent elements. Thus, the molar mass of \( \mathrm{WO}_3 \) is 231.8 g/mol.
• Conversion Between Moles and Grams: To convert mass to moles, divide by the molar mass. Conversely, multiply moles by the molar mass to find the mass in grams.

Molar mass calculations are essential for determining the quantities of materials needed or produced in a chemical reaction.

Balanced Equations

Balancing a chemical equation ensures that the same number of atoms of each element are present on both sides of the equation. This reflects the conservation of mass in a chemical reaction.

• Why Balance Equations? It ensures the correct stoichiometric relationships between reactants and products, which is vital for any reaction calculations.
• The Process: Involves adjusting the coefficients before formulas to make sure there is an equal number of each type of atom on both sides. This does not change the substances involved, only the proportions.
• The Balanced Reaction for Tungsten Production: The balanced equation \( \mathrm{WO}_3(s) + 3 \mathrm{H}_2(g) \rightarrow \mathrm{W}(s) + 3 \mathrm{H}_2 \mathrm{O}(g) \) indicates that 1 mole of tungsten(VI) oxide reacts with 3 moles of hydrogen to produce 1 mole of tungsten and 3 moles of water vapor.

Balanced equations are the foundation for calculating how much of each substance is needed or formed in a chemical reaction.