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Tungsten metal, W, is used to make incandescent bulb filaments. The metal is produced from the yellow tungsten(VI) oxide, \(\mathrm{WO}_{3}\), by reaction with hydrogen. $$ \mathrm{WO}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{W}(s)+3 \mathrm{H}_{2} \mathrm{O}(g) $$ How many grams of tungsten can be obtained from \(4.81 \mathrm{~kg}\) of hydrogen with excess tungsten(VI) oxide?

Short Answer

Expert verified
145.9027 kg of tungsten can be obtained.

Step by step solution

01

Write the Balanced Chemical Equation

The balanced chemical equation for the reaction is already provided: \( \mathrm{WO}_{3}(s)+3 \mathrm{H}_{2}(g) \rightarrow \mathrm{W}(s)+3 \mathrm{H}_{2} \mathrm{O}(g) \). This tells us that 3 moles of hydrogen gas \((\mathrm{H}_2)\) react with 1 mole of \(\mathrm{WO}_3\) to produce 1 mole of tungsten \((\mathrm{W})\) and 3 moles of water vapor \((\mathrm{H}_2\mathrm{O})\).
02

Calculate Moles of Hydrogen

First, calculate the number of moles of hydrogen \((\mathrm{H}_2)\). The molar mass of hydrogen gas \((\mathrm{H}_2)\) is 2.02 g/mol. Therefore, the number of moles of \(\mathrm{H}_2\) in 4.81 kg is \( \frac{4810 \text{ g}}{2.02 \text{ g/mol}} \approx 2380.2 \text{ moles}.\)
03

Use Stoichiometry to Find Moles of Tungsten

From the balanced equation, 3 moles of \(\mathrm{H}_2\) produce 1 mole of \(\mathrm{W}\). So, \( \frac{2380.2}{3} \approx 793.4 \) moles of \(\mathrm{W}\) can be formed from 2380.2 moles of hydrogen.
04

Calculate Grams of Tungsten

The molar mass of tungsten \((\mathrm{W})\) is 183.84 g/mol. Multiply the moles of tungsten by its molar mass to get the mass: \( 793.4 \text{ moles} \times 183.84 \text{ g/mol} \approx 145902.7 \text{ g} \). Convert grams to kilograms for the final answer: \(145902.7 \text{ g} = 145.9027 \text{ kg}\).
05

Conclusion

Therefore, under the given conditions, \(145.9027 \text{ kg} \) of tungsten can be obtained from \(4.81 \text{ kg} \) of hydrogen with excess \(\mathrm{WO}_3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tungsten
Tungsten, known by its chemical symbol "W," is a remarkable metal. It's highly valued for its unique properties, making it perfect for specific industrial applications. For instance, its impressive melting point and thermal conductivity make tungsten ideal for crafting incandescent bulb filaments. Tungsten's dense nature is useful in various fields, including military applications, electronics, and high-temperature environments.
Tungsten's primary source is tungsten(VI) oxide, \(\mathrm{WO}_{3}\), which is reduced using hydrogen, \(\mathrm{H}_{2}\), to produce metallic tungsten, \(\mathrm{W}\). This reaction plays an essential role in obtaining metallic tungsten, demonstrating the effectiveness of tungsten's extraction from its oxide form.
Balanced Chemical Equation
In chemistry, understanding reactions starts with a balanced chemical equation. This equation showcases how reactants transform into products, ensuring that the number of atoms for each element is conserved.
The reaction to produce tungsten from tungsten(VI) oxide and hydrogen is given by:\[ \mathrm{WO}_{3}(s)+3 \mathrm{H}_{2}(g) \rightarrow \mathrm{W}(s)+3 \mathrm{H}_{2}\mathrm{O}(g) \]
This equation tells us that one mole of tungsten(VI) oxide reacts with three moles of hydrogen gas to produce one mole of tungsten and three moles of water vapor. Balancing equations is crucial to stoichiometry as it provides the exact proportions of reactants and products involved, necessary for calculations like determining the mass of products formed.
Molar Mass
Molar mass is a fundamental concept in stoichiometry, helping chemists relate moles to grams through atomic masses. The molar mass of a substance represents the mass of one mole of its particles, typically measured in grams per mole (g/mol).
For instance, the molar mass of hydrogen (\(\mathrm{H}_2\)) is 2.02 g/mol, and tungsten (\(\mathrm{W}\)) has a molar mass of 183.84 g/mol. Knowing these values, we can deduce the amount of substance in terms of mass for any given mole quantity.
This concept is critical in calculations where you're converting between mass and moles, as shown when determining how much tungsten is producible from a given mass of hydrogen in the reaction discussed.
Hydrogen
Hydrogen is the lightest and most abundant element in the universe. It exists predominantly as a diatomic molecule (\(\mathrm{H}_2\)). Hydrogen serves various roles in chemical reactions, especially as a reducing agent. This property is evident in its use to convert tungsten(VI) oxide into metallic tungsten.
In the given reaction, hydrogen gas reacts with tungsten(VI) oxide to yield tungsten and water vapor. The balanced equation shows that three moles of hydrogen are necessary to reduce one mole of tungsten(VI) oxide fully. Understanding the role of hydrogen in these reactions helps in comprehending stoichiometric calculations essential for predicting the quantities of materials produced or consumed in reactions.

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Most popular questions from this chapter

Part 1 a. How many hydrogen and oxygen atoms are present in 1 molecule of \(\mathrm{H}_{2} \mathrm{O} ?\) b. How many moles of hydrogen and oxygen atoms are present in \(1 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) ? c. What are the masses of hydrogen and oxygen in \(1.0 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) ? d. What is the mass of \(1.0 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) ? Part 2: Two hypothetical ionic compounds are discovered with the chemical formulas \(\mathrm{XCl}_{2}\) and \(\mathrm{YCl}_{2}\), where \(\mathrm{X}\) and \(\mathrm{Y}\) represent symbols of the imaginary elements. Chemical analysis of the two compounds reveals that \(0.25 \mathrm{~mol} \mathrm{XCl}_{2}\) has a mass of \(100.0 \mathrm{~g}\) and \(0.50 \mathrm{~mol} \mathrm{YCl}_{2}\) has a mass of \(125.0 \mathrm{~g}\). a. What are the molar masses of \(\mathrm{XCl}_{2}\) and \(\mathrm{YCl}_{2}\) ? b. If you had \(1.0\) -mol samples of \(\mathrm{XCl}_{2}\) and \(\mathrm{YCl}_{2}\), how would the number of chloride ions compare? C. If you had \(1.0\) -mol samples of \(\mathrm{XCl}_{2}\) and \(\mathrm{YCl}_{2}\), how would the masses of elements \(\mathrm{X}\) and \(\mathrm{Y}\) compare? d. What is the mass of chloride ions present in \(1.0 \mathrm{~mol} \mathrm{XCl}_{2}\) and \(1.0 \mathrm{~mol} \mathrm{YCl}_{2} ?\) e. What are the molar masses of elements \(\mathrm{X}\) and \(\mathrm{Y}\) ? f. How many moles of \(\mathrm{X}\) ions and chloride ions would be present in a \(200.0-\mathrm{g}\) sample of \(\mathrm{XCl}_{2}\) ? g. How many grams of \(Y\) ions would be present in a \(250.0-\mathrm{g}\) sample of \(\mathrm{YCl}_{2} ?\) h. What would be the molar mass of the compound \(\mathrm{YBr}_{3}\) ? Part 3: A minute sample of \(\mathrm{AlCl}_{3}\) is analyzed for chlorine. The analysis reveals that there are 12 chloride ions present in the sample. How many aluminum ions must be present in the sample? a. What is the total mass of \(\mathrm{AlCl}_{3}\) in this sample? b. How many moles of \(\mathrm{AlCl}_{3}\) are in this sample?

Hydrogen cyanide, HCN, can be made by a two-step process. First, ammonia is reacted with \(\mathrm{O}_{2}\) to give nitric oxide, \(\mathrm{NO}\). $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ Then nitric oxide is reacted with methane, \(\mathrm{CH}_{4}\). $$ 2 \mathrm{NO}(g)+2 \mathrm{CH}_{4}(g) \longrightarrow 2 \mathrm{HCN}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{H}_{2}(g) $$ When \(24.2 \mathrm{~g}\) of ammonia and \(25.1 \mathrm{~g}\) of methane are used, how many grams of hydrogen cyanide can be produced?

Methyl salicylate (oil of wintergreen) is prepared by heating salicylic acid, \(\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}\), with methanol, \(\mathrm{CH}_{3} \mathrm{OH}\). $$ \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}+\mathrm{CH}_{3} \mathrm{OH} \longrightarrow \mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}_{3}+\mathrm{H}_{2} \mathrm{O} $$ In an experiment, \(1.50 \mathrm{~g}\) of salicylic acid is reacted with \(11.20 \mathrm{~g}\) of methanol. The yield of methyl salicylate, \(\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}_{3}\), is \(1.27 \mathrm{~g}\). What is the percentage yield?

A 3.41-g sample of a metallic element, M, reacts completely with \(0.0158\) mol of a gas, \(\mathrm{X}_{2}\), to form \(4.52\) g MX. What are the identities of \(\mathrm{M}\) and \(\mathrm{X}\) ?

Exactly \(4.0 \mathrm{~g}\) of hydrogen gas combines with \(32 \mathrm{~g}\) of oxygen gas according to the following reaction. $$ 2 \mathrm{H}_{2}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} $$ a. How many hydrogen molecules are required to completely react with 48 oxygen molecules? b. If you ran the reaction and it produced \(5.0 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\), how many moles of both \(\mathrm{O}_{2}\) and \(\mathrm{H}_{2}\) did you start with? c. If you started with \(37.5 \mathrm{~g} \mathrm{O}_{2}\), how many grams of \(\mathrm{H}_{2}\) did you start with to have a complete reaction? d. How many grams of \(\mathrm{O}_{2}\) and \(\mathrm{H}_{2}\) were reacted to produce \(30.0 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} ?\)

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