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Chapter 3: Problem 13

Hew-many grams of \(\mathrm{NH}_{3}\) will have the same number of molecules as \(15.0 \mathrm{~g}\) of \(\mathrm{C}_{6} \mathrm{H}_{6} ?\) \(\begin{array}{ll}3.27 & 1.92\end{array}\) $$ 15.0 $$ \(17.0 \quad 14.2\)

Short Answer

Expert verified
3.27 grams of \(\mathrm{NH}_{3}\).

Step by step solution

01

Calculate Moles of Benzene

First, calculate the number of moles in 15.0 g of benzene (\(\mathrm{C}_{6}\mathrm{H}_{6}\)). Find the molar mass of benzene by adding the atomic masses of its atoms. \(\mathrm{C} = 12.01 \; \mathrm{g/mol}, \text{and}\ \mathrm{H} = 1.008 \; \mathrm{g/mol}\). The molar mass of \(\mathrm{C}_{6} \mathrm{H}_{6}\) is: \[6(12.01\; \mathrm{g/mol}) + 6(1.008\; \mathrm{g/mol}) = 78.11\; \mathrm{g/mol}\]Now, calculate the moles of \(\mathrm{C}_{6}\mathrm{H}_{6}\):\[\frac{15.0 \; \mathrm{g}}{78.11 \; \mathrm{g/mol}} \approx 0.192 \; \mathrm{mol} \]
02

Calculate Mole Quantity for Ammonia

Since we need the same number of molecules as benzene, we find the equivalent moles of ammonia. The number of moles of \(\mathrm{NH}_{3}\) needed is:\[0.192 \; \mathrm{mol} \]We calculated this from the moles of benzene, assuming an equal number of molecules is wanted.
03

Determine Molar Mass of Ammonia

The molar mass of \(\mathrm{NH}_{3}\) needs to be found to convert from moles to grams. Calculate it by adding the atomic masses: \[\mathrm{N} = 14.01 \; \mathrm{g/mol}, \quad \mathrm{H} = 1.008 \; \mathrm{g/mol}\] Then, the molar mass is: \[14.01 \; \mathrm{g/mol} + 3 \times 1.008 \; \mathrm{g/mol} = 17.034 \; \mathrm{g/mol}\]
04

Convert Moles of Ammonia to Grams

Multiply the moles of \(\mathrm{NH}_{3}\) by its molar mass to find the mass in grams:\[0.192 \; \mathrm{mol} \times 17.034 \; \mathrm{g/mol} \approx 3.27 \; \mathrm{g}\]
05

Conclusion

Therefore, approximately 3.27 grams of \(\mathrm{NH}_{3}\) have the same number of molecules as 15.0 grams of \(\mathrm{C}_{6}\mathrm{H}_{6}\).

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Key Concepts

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

Molecular Equivalence
Molecular equivalence is an important concept in chemistry. It means comparing two substances to find a common ground based on the number of molecules.
In this exercise, we need to ascertain how many grams of one substance equals the same number of molecules as a different substance.
The law of Avogadro tells us that one mole of any substance contains the same number of molecules, which is Avogadro's number: approximately \(6.022 \times 10^{23}\) molecules.
  • In the exercise, both benzene (C_6H_6) and ammonia (NH_3) need to have the same number of molecules.
  • This means if we know how many moles of benzene are in 15 g, the same moles of ammonia will have an equivalent number of molecules.
Understanding molecular equivalence allows one to transition smoothly between different substances by keeping the comparison on a molecular level.
Molar Calculations
Molar calculations are vital for finding the amount of substance when dealing with molecules. Here's how they play a role in this type of problem:
To begin this calculation, we need the molar mass of benzene, calculated as follows:
  • The molar mass of benzene: 6 carbon atoms \(6 \times 12.01 \; \mathrm{g/mol}\) plus 6 hydrogen atoms \(6 \times 1.008 \; \mathrm{g/mol}\), totaling \(78.11 \; \mathrm{g/mol}\).
  • Then, to determine the moles of benzene in 15 g, use the formula for moles: \(\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}\).
  • Substitute into the formula: \(\frac{15.0 \; \mathrm{g}}{78.11 \; \mathrm{g/mol}} = 0.192 \; \mathrm{mol}\).
The process is similar for ammonia with its own molar mass calculation, enabling the comparison of the two by their molecular content.
Mass Conversion in Chemistry
Mass conversion in chemistry involves translating between the size of a substance in moles to its mass in grams. This conversion hinges on molar mass.
  • For ammonia (NH_3), the molar mass is computed as \(14.01 \; \mathrm{g/mol}\) for nitrogen plus \(3 \times 1.008 \; \mathrm{g/mol}\) for hydrogen, resulting in \(17.034 \; \mathrm{g/mol}\).
  • To find out how much ammonia weighs that would be equivalent to the moles of benzene, the number of moles calculated prior is multiplied by ammonia's molar mass. That is, \(0.192 \; \mathrm{mol} \times 17.034 \; \mathrm{g/mol} \approx 3.27 \; \mathrm{g}\).
Thus, mass conversion may start from molecules to moles before finding grams, balancing equations, and reallocating resources to ensure equal comparisons between different chemical substances.

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Most popular questions from this chapter

Find the formula weights of the following substances to three significant figures. sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) phosphorus pentachloride, \(\mathrm{PCl}_{5}\) ammonium chloride, \(\mathrm{NH}_{4} \mathrm{Cl}\) calcium hydroxide, \(\mathrm{Ca}(\mathrm{OH})_{2}\)

Hydrogen cyanide, \(\mathrm{HCN},\) can be made by a two-step process. First, ammonia is reacted with \(\mathrm{O}_{2}\) to give nitric oxide, 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?

Calcium sulfate, \(\mathrm{CaSO}_{4}\), is a white, crystalline powder. Gypsum is a mineral, or natural substance, that is a hydrate of calcium sulfate. A 1.000 -g sample of gypsum contains \(0.791 \mathrm{~g} \mathrm{CaSO}_{4} .\) How many moles of \(\mathrm{CaSO}_{4}\) are there in this sample? Assuming that the rest of the sample is water, how many moles of \(\mathrm{H}_{2} \mathrm{O}\) are there in the sample? Show that the result is consistent with the formula \(\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O} .\)

When dinitrogen pentoxide, \(\mathrm{N}_{2} \mathrm{O}_{5},\) a white solid, is heated, it decomposes to nitrogen dioxide and oxygen. $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5}(s) \stackrel{\Delta}{\longrightarrow} 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ If a sample of \(\mathrm{N}_{2} \mathrm{O}_{5}\) produces \(1.381 \mathrm{~g} \mathrm{O}_{2}\), how many grams of \(\mathrm{NO}_{2}\) are formed?

Alloys, or metallic mixtures, of mercury with another metal are called amalgams. Sodium in sodium amalgam reacts with water. (Mercury does not.) $$ 2 \mathrm{Na}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g) $$ If a \(15.23-\mathrm{g}\) sample of sodium amalgam evolves \(0.108 \mathrm{~g}\) of hydrogen, what is the percentage of sodium in the amalgam?
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