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How many grams of \(\mathrm{NH}_{3}\) will have the same number of molecules as \(15.0 \mathrm{~g}\) of \(\mathrm{C}_{6} \mathrm{H}_{6} ?\) a. \(3.27\) b. \(1.92\) c. \(15.0\) d. \(17.0\) e. \(14.2\)

Short Answer

Expert verified
The answer is 3.27 grams of NH3.

Step by step solution

01

Determine Molar Mass of C6H6

Calculate the molar mass of benzene, \( C_6H_6 \). It consists of 6 carbon atoms and 6 hydrogen atoms. The molar mass is given by:\[(6 \times 12.01 \, \text{g/mol}) + (6 \times 1.01 \, \text{g/mol}) = 78.12 \, \text{g/mol}.\]
02

Calculate Moles of C6H6

Use the molar mass to find the number of moles of benzene in 15.0 g using the formula:\[\text{moles of } C_6H_6 = \frac{15.0 \, \text{g}}{78.12 \, \text{g/mol}} \approx 0.192 \, \text{mol}.\]
03

Determine Moles of NH3

Since the number of molecules in \( C_6H_6 \) must equal the number of molecules in \( NH_3 \), the number of moles of \( NH_3 \) must also be 0.192 mol.
04

Determine Molar Mass of NH3

Calculate the molar mass of ammonia, \( NH_3 \), which consists of 1 nitrogen and 3 hydrogen atoms:\[(1 \times 14.01 \, \text{g/mol}) + (3 \times 1.01 \, \text{g/mol}) = 17.04 \, \text{g/mol}.\]
05

Calculate Grams of NH3

Use the moles of \( NH_3 \) and its molar mass to find the number of grams:\[\text{mass of } NH_3 = 0.192 \, \text{mol} \times 17.04 \, \text{g/mol} \approx 3.27 \, \text{g}.\]
06

Conclusion

Through the above calculations, we find that 3.27 grams of \( NH_3 \) contain the same number of molecules as 15.0 grams of \( C_6H_6 \).

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Key Concepts

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

Molar Mass Calculation
The concept of molar mass is essential in chemistry for converting between grams and moles. Molar mass, also known as molecular weight, is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It allows you to compare different substances based on their mass at the molecular level. To calculate the molar mass, you simply sum the atomic masses of all the atoms in a molecule, which can be found on the periodic table. For example, in calculating the molar mass of benzene, - Count the number of each type of atom: 6 carbon atoms and 6 hydrogen atoms.
- Multiply the number of each atom by its atomic mass: - Carbon: 6 atoms × 12.01 g/mol = 72.06 g/mol - Hydrogen: 6 atoms × 1.01 g/mol = 6.06 g/mol - Add them together to get a total molar mass of 78.12 g/mol for benzene. Similarly, for ammonia, - Count the atoms: 1 nitrogen and 3 hydrogen atoms. - Calculate: - Nitrogen: 1 atom × 14.01 g/mol = 14.01 g/mol - Hydrogen: 3 atoms × 1.01 g/mol = 3.03 g/mol - This results in a total molar mass of 17.04 g/mol for ammonia.
Chemical Calculations
Chemical calculations involve using various mathematical techniques to determine quantities of substances. These calculations are often based on the principle of moles, which relate to Avogadro's number, approximately \[6.022 \times 10^{23}\] molecules in one mole.When performing chemical calculations, such as in this exercise, it is crucial to convert mass to moles using the molar mass. For example, when given 15.0 grams of benzene, the number of moles can be found by dividing its mass by its molar mass:\[\text{moles of benzene} = \frac{15.0 \, \text{g}}{78.12 \, \text{g/mol}} \approx 0.192 \, \text{mol}.\]This process can then be applied to other substances in the reaction. In the case of ammonia, the same number of moles will have the same number of molecules, showing the importance of moles in balancing chemical reactions accurately and simply.
Stoichiometry
Stoichiometry is a branch of chemistry that focuses on the quantitative relationships between reactants and products in a chemical reaction. It enables chemists to predict the outcomes of reactions by using balanced equations and involves key concepts like molar ratios and conversions between different units. Stochiometry relies on the fact that atoms are conserved in reactions, meaning the number of atoms of each element remains constant. In this exercise involving benzene and ammonia, - Ensure the number of molecules is the same by equating the moles. - By determined calculations, where both benzene and ammonia share an equal 0.192 moles, the stoichiometric balance asserts their molecule count is identical. In summary: - Stoichiometry is a key tool for chemists to manage and predict the changes in chemical reactions.
- It combines the principles of moles, conservation of mass, and balanced equations to solve complex chemical problems efficiently.

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Most popular questions from this chapter

Phosphorus oxychloride is the starting compound for preparing substances used as flame retardants for plastics. An \(8.53\) -mg sample of phosphorus oxychloride contains \(1.72 \mathrm{mg}\) of phosphorus. What is the mass percentage of phosphorus in the compound?

Calcium carbide, \(\mathrm{CaC}_{2}\), used to produce acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}\), is prepared by heating calcium oxide, \(\mathrm{CaO}\), and carbon, \(\mathrm{C}\), to high temperature. $$ \mathrm{CaO}(s)+3 \mathrm{C}(s) \longrightarrow \mathrm{CaC}_{2}(s)+\mathrm{CO}(g) $$ If a mixture contains \(2.60 \mathrm{~kg}\) of each reactant, how many grams of calcium carbide can be prepared?

While cleaning out your closet, you find a jar labeled "2.21 moles lead nitrite." Since Stock convention was not used, you do not know the oxidation number of the lead. You weigh the contents, and find a mass of \(6.61 \times 10^{5} \mathrm{mg} .\) What is the percentage composition of nitrite?

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 \(\mathrm{X}\) ?

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?

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