For the reaction shown, calculate how many moles of each product form when the given amount of each reactant completely reacts. Assume there is more than enough of the other reactant. $$ \mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) \(4.6 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8}\) (b) \(4.6 \mathrm{~mol} \mathrm{O}_{2}\) (c) \(0.0558 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8}\) (d) \(0.0558 \mathrm{~mol} \mathrm{O}_{2}\)

Short Answer

Expert verified
For 4.6 mol \(\mathrm{C}_{3}\mathrm{H}_{8}\), 13.8 mol \(\mathrm{CO}_{2}\) and 18.4 mol \(\mathrm{H}_{2} \mathrm{O}\) form. For 4.6 mol \(\mathrm{O}_{2}\), 2.76 mol \(\mathrm{CO}_{2}\) and 3.68 mol \(\mathrm{H}_{2} \mathrm{O}\) form. For 0.0558 mol \(\mathrm{C}_{3}\mathrm{H}_{8}\), 0.1674 mol \(\mathrm{CO}_{2}\) and 0.2232 mol \(\mathrm{H}_{2} \mathrm{O}\) form. For 0.0558 mol \(\mathrm{O}_{2}\), 0.03348 mol \(\mathrm{CO}_{2}\) and 0.04464 mol \(\mathrm{H}_{2} \mathrm{O}\) form.

Step by step solution

01

Understand the Balanced Reaction

The balanced chemical reaction is \[ \mathrm{C}_{3} \mathrm{H}_{8}(g) + 5 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g) + 4 \mathrm{H}_{2} \mathrm{O}(g) \]This indicates that 1 mole of propane (\(\mathrm{C}_{3}\mathrm{H}_{8}\)) reacts with 5 moles of oxygen (\(\mathrm{O}_{2}\)) to produce 3 moles of carbon dioxide (\(\mathrm{CO}_{2}\)) and 4 moles of water (\(\mathrm{H}_{2} \mathrm{O}\)).
02

Moles of Products from 4.6 mol C3H8

According to the stoichiometry of the reaction, 1 mole of \(\mathrm{C}_{3}\mathrm{H}_{8}\) yields 3 moles of \(\mathrm{CO}_{2}\) and 4 moles of \(\mathrm{H}_{2} \mathrm{O}\). Therefore, for 4.6 moles of \(\mathrm{C}_{3}\mathrm{H}_{8}\):- Moles of \(\mathrm{CO}_{2}\) formed: \[4.6 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8} \times \frac{3 \mathrm{~mol} \mathrm{CO}_{2}}{1 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8}} = 13.8 \mathrm{~mol} \mathrm{CO}_{2}\]- Moles of \(\mathrm{H}_{2} \mathrm{O}\) formed:\[4.6 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8} \times \frac{4 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8}} = 18.4 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\]
03

Moles of Products from 4.6 mol O2

Based on the stoichiometry, 5 moles of \(\mathrm{O}_{2}\) react to produce 3 moles of \(\mathrm{CO}_{2}\) and 4 moles of \(\mathrm{H}_{2} \mathrm{O}\). Given 4.6 moles of \(\mathrm{O}_{2}\), we divide by 5 to find the equivalent moles of \(\mathrm{C}_{3}\mathrm{H}_{8}\), and then calculate products:- Moles of \(\mathrm{CO}_{2}\):\[4.6 \mathrm{~mol} \mathrm{O}_{2} \times \frac{1 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8}}{5 \mathrm{~mol} \mathrm{O}_{2}} \times \frac{3 \mathrm{~mol} \mathrm{CO}_{2}}{1 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8}} = 2.76 \mathrm{~mol} \mathrm{CO}_{2}\]- Moles of \(\mathrm{H}_{2} \mathrm{O}\):\[4.6 \mathrm{~mol} \mathrm{O}_{2} \times \frac{1 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8}}{5 \mathrm{~mol} \mathrm{O}_{2}} \times \frac{4 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8}} = 3.68 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\]
04

Moles of Products from 0.0558 mol C3H8

Using the stoichiometry ratio, we can calculate products from 0.0558 moles of \(\mathrm{C}_{3}\mathrm{H}_{8}\):- Moles of \(\mathrm{CO}_{2}\):\[0.0558 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8} \times \frac{3 \mathrm{~mol} \mathrm{CO}_{2}}{1 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8}} = 0.1674 \mathrm{~mol} \mathrm{CO}_{2}\]- Moles of \(\mathrm{H}_{2} \mathrm{O}\):\[0.0558 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8} \times \frac{4 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8}} =0.2232 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\]
05

Moles of Products from 0.0558 mol O2

For 0.0558 moles of \(\mathrm{O}_{2}\), calculate corresponding moles of \(\mathrm{C}_{3}\mathrm{H}_{8}\), and then moles of products as before:- Moles of \(\mathrm{CO}_{2}\):\[0.0558 \mathrm{~mol} \mathrm{O}_{2} \times \frac{1 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8}}{5 \mathrm{~mol} \mathrm{O}_{2}} \times \frac{3 \mathrm{~mol} \mathrm{CO}_{2}}{1 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8}} = 0.03348 \mathrm{~mol} \mathrm{CO}_{2}\]- Moles of \(\mathrm{H}_{2} \mathrm{O}\):\[0.0558 \mathrm{~mol} \mathrm{O}_{2} \times \frac{1 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8}}{5 \mathrm{~mol} \mathrm{O}_{2}} \times \frac{4 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8}} = 0.04464 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\]

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Key Concepts

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

Chemical Reaction Balancing
When it comes to understanding chemical reactions, the first step is balancing the equation. A balanced chemical equation ensures that the same number of atoms of each element is present on both sides of the equation, a principle known as the conservation of mass.

For instance, in the reaction \(\mathrm{C}_{3}\mathrm{H}_{8}(g) + 5\mathrm{O}_{2}(g) \longrightarrow 3\mathrm{CO}_{2}(g) + 4\mathrm{H}_{2}\mathrm{O}(g)\), we see that each carbon (C), hydrogen (H), and oxygen (O) atom on the left is accounted for on the right. Precisely, one molecule of propane (\(\mathrm{C}_{3}\mathrm{H}_{8}\)) contains 3 carbon atoms and 8 hydrogen atoms, while the 5 molecules of oxygen (\(\mathrm{O}_{2}\)) provide a total of 10 oxygen atoms.

On the product side, there are 3 molecules of carbon dioxide (\(\mathrm{CO}_{2}\)) and 4 molecules of water (\(\mathrm{H}_{2}\mathrm{O}\)), ensuring that we have the same tally of atoms: 3 carbon, 8 hydrogen, and 10 oxygen. It's this balance that allows us to accurately predict the amounts of products formed from given reactants, which is the heart of stoichiometric calculations.
Mole Concept
At the core of chemistry lies the mole concept, which is simply a way to count atoms and molecules by weighing them. Essentially, one mole is equal to Avogadro's number (approximately \(6.022 \times 10^{23}\) entities) of atoms or molecules. This is immensely practical since we cannot count tiny atoms individually, but we can weigh out moles of substances.

In the given exercise, we encounter terms like '4.6 moles of \(\mathrm{C}_{3}\mathrm{H}_{8}\)', which means we have 4.6 times Avogadro's number of \(\mathrm{C}_{3}\mathrm{H}_{8}\) molecules. Whether we are dealing with a small sample of 0.0558 moles or a larger amount, the mole allows us to use mass as a proxy for counting out exact numbers of atoms and molecules for our reactions.

Understanding the mole concept is crucial for stoichiometry because it allows us to convert between mass and number of particles, and thereby to calculate the quantities of reactants consumed and products formed in a chemical reaction.
Stoichiometric Calculations
Stoichiometry is the aspect of chemistry that involves calculating the quantities of reactants and products in chemical reactions. It relies on the balanced chemical equation and the mole concept mentioned earlier. The ratios of the substances in a balanced chemical equation tell us the ratios in which substances react and are produced; these ratios are the stoichiometric coefficients.

For example, the reaction from our exercise shows that for each mole of propane (\(\mathrm{C}_{3}\mathrm{H}_{8}\)), 3 moles of carbon dioxide (\(\mathrm{CO}_{2}\)) and 4 moles of water (\(\mathrm{H}_{2}\mathrm{O}\)) are produced. This means if we start with any number of moles of \(\mathrm{C}_{3}\mathrm{H}_{8}\), we can multiply it by 3 to find the moles of \(\mathrm{CO}_{2}\) produced or by 4 to determine the moles of \(\mathrm{H}_{2}\mathrm{O}\) produced. This is exactly what's done in the step-by-step solution to the given problem.

Understanding stoichiometric calculations is vital to predicting product formation and is invaluable in fields such as pharmaceuticals, material science, and any situation where precise chemical composition is critical.

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Most popular questions from this chapter

Consider the unbalanced equation for the reaction of aluminum with sulfuric acid: $$ \mathrm{Al}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2}(g) $$ (a) Balance the equation. (b) How many moles of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) are required to completely react with \(8.3 \mathrm{~mol}\) of \(\mathrm{Al}\) ? (c) How many moles of \(\mathrm{H}_{2}\) are formed by the complete reaction of \(0.341 \mathrm{~mol}\) of \(\mathrm{Al}\) ?

Consider the reaction between reactants \(\mathrm{S}\) and \(\mathrm{O}_{2}\) : $$ 2 \mathrm{~S}(\mathrm{~s})+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g) $$ If a reaction vessel initially contains \(5 \mathrm{~mol} \mathrm{~S}\) and \(9 \mathrm{~mol} \mathrm{O}_{2}\), how many moles of \(\mathrm{S}, \mathrm{O}_{2}\), and \(\mathrm{SO}_{3}\) will be in the reaction vessel after the reactants have reacted as much as possible? (Assume \(100 \%\) actual yield.)

Consider the unbalanced equation for the neutralization of acetic acid: $$ \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Ca}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q) $$ Balance the equation and determine how many moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) are required to completely neutralize \(1.07 \mathrm{~mol}\) of \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\).

Explain the relationship between the sign of \(\Delta H_{\mathrm{rxn}}\) and whether a reaction is exothermic or endothermic.

For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts. $$ 2 \mathrm{HgO}(\mathrm{s}) \longrightarrow 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g) $$ (a) \(2.13 \mathrm{~g} \mathrm{Hg} \mathrm{O}\) (b) \(6.77 \mathrm{~g} \mathrm{Hg} \mathrm{O}\) (c) \(1.55 \mathrm{~kg} \mathrm{Hg} \mathrm{O}\) (d) \(3.87 \mathrm{mg} \mathrm{HgO}\)

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