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(a) What volumes (in liters) of ammonia and oxygen must react to form \(12.8 \mathrm{~L}\) of nitric oxide according to the equation at the same temperature and pressure? $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ (b) What volumes (in liters) of propane and water vapor must react to form \(8.96 \mathrm{~L}\) of hydrogen according to the equation at the same temperature and pressure? $$ \mathrm{C}_{3} \mathrm{H}_{8}(g)+3 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 3 \mathrm{CO}(g)+7 \mathrm{H}_{2}(g) $$

Short Answer

Expert verified
For part (a), 12.8 L of \(NH_{3}\) and 16 L of \(O_{2}\) are required. For part (b), 1.254 L of \(C_{3}H_{8}\) and 3.853 L of \(H_{2}O\) are required.

Step by step solution

01

Establish Mole Ratios for the First Reaction

From the balanced chemical equation \(4NH_{3}(g) + 5O_{2}(g) \rightarrow 4NO(g) + 6H_{2}O(g)\), we can establish that 4 volumes of \(NH_{3}\), and 5 volumes of \(O_{2}\), react to form 4 volumes of \(NO\). This means ratio of the volume of \(NH_{3}\) to the volume of \(NO\) is 1 (i.e., 4/4), and the ratio of the volume of \(O_{2}\) to the volume of \(NO\) is 1.25 (i.e., 5/4).
02

Calculate Volumes for the First Reaction

We know \(12.8 L\) of \(NO\) are formed. So using the volume ratios established in step 1, the volume of \(NH_{3}\) required would be \(12.8 L * 1 = 12.8 L\), and the volume of \(O_{2}\) required would be \(12.8 L * 1.25 = 16 L\).
03

Establish Mole Ratios for the Second Reaction

From the balanced chemical equation \(C_{3}H_{8}(g) + 3H_{2}O(g) \rightarrow 3CO(g) + 7H_{2}(g)\), we can establish that 1 volume of \(C_{3}H_{8}\), and 3 volumes of \(H_{2}O\), react to form 7 volumes of \(H_{2}\). This means the ratio of the volume of \(C_{3}H_{8}\) to the volume of \(H_{2}\) is 0.14 (i.e., 1/7), and the ratio of the volume of \(H_{2}O\) to the volume of \(H_{2}\) is 0.43 (i.e., 3/7).
04

Calculate Volumes for the Second Reaction

We know \(8.96 L\) of \(H_{2}\) are formed. So using the volume ratios established in step 3, the volume of \(C_{3}H_{8}\) required would be \(8.96 L * 0.14 = 1.254 L\), and the volume of \(H_{2}O\) required would be \(8.96 L * 0.43 = 3.853 L\).

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Key Concepts

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

Understanding Volume Ratios in Reactions
In chemistry, particularly in stoichiometry, volume ratios help us understand how much of each reactant is needed in a gaseous chemical reaction, or how much product we can form.Volume ratios come from the coefficients of the balanced equation, which illustrate the number of moles needed for the reaction. In gases, due to Avogadro's law, these mole ratios translate directly into volume ratios when the gases are at the same temperature and pressure.For instance, in the reaction between ammonia (\(\mathrm{NH}_3\)) and oxygen (\(\mathrm{O}_2\)) to produce nitric oxide (\(\mathrm{NO}\)), the equation says:{linebreak}\[4\,\mathrm{NH}_3(g) + 5\,\mathrm{O}_2(g) \rightarrow 4\,\mathrm{NO}(g) + 6\,\mathrm{H}_2\mathrm{O}(g)\]{linebreak}We deduce that:
  • 4 volumes of \(\mathrm{NH}_3\) react with 5 volumes of \(\mathrm{O}_2\)
  • This forms 4 volumes of \(\mathrm{NO}\)
This gives a straightforward volume ratio of 1:1 for \(\mathrm{NH}_3\) to \(\mathrm{NO}\), and a ratio of 1.25:1 for \(\mathrm{O}_2\) to \(\mathrm{NO}\) because 5 divided by 4 equals 1.25.
Basics of Chemical Reactions
Chemical reactions involve the transformation of substances through breaking and forming chemical bonds. In a balanced chemical equation, the number of each type of atom is the same on both sides of the equation.The reaction of propane (\(\mathrm{C}_3\mathrm{H}_8\)) with steam (\(\mathrm{H}_2\mathrm{O}\)) demonstrates stoichiometric principles. The balanced equation is:{linebreak}\[\mathrm{C}_3\mathrm{H}_8(g) + 3\,\mathrm{H}_2\mathrm{O}(g) \rightarrow 3\,\mathrm{CO}(g) + 7\,\mathrm{H}_2(g)\]{linebreak}Here’s how we interpret this:
  • 1 molecule (or unit) of \(\mathrm{C}_3\mathrm{H}_8\) reacts with 3 molecules of \(\mathrm{H}_2\mathrm{O}\)
  • This forms 3 molecules of \(\mathrm{CO}\) and 7 molecules of \(\mathrm{H}_2\)
Understanding these numbers shows us that the ratios in our equation dictate how much of each reactant is needed, and how much product will be formed.Balanced equations allow us to predict and calculate these amounts accurately, showcasing the law of conservation of mass.
Explaining Gas Laws in Chemistry
Gas laws are fundamental in stoichiometric calculations involving gaseous substances. They describe the behavior of gases in relation to temperature, volume, and pressure. One of the key laws is Avogadro's Law, which states that "equal volumes of gases at the same temperature and pressure contain the same number of molecules." This principle allows mole ratios in chemical equations to be treated as volume ratios when gases are involved. In stoichiometry, when identities of gases and their conditions remain constant, we use these relationships:
  • Given the volumes of gases at a uniform temperature and pressure, the coefficients in a balanced chemical equation directly translate into volume ratios.
  • Boyle’s Law and Charles’s Law further guide how changing conditions would affect gas volumes.
For activities at constant conditions, the simplifying assumption often used is that the volumes of reactants and products align exactly as the coefficients in a balanced equation suggest. Through applying these laws, complex calculations become more manageable, and we gain better insight into how changes in conditions affect reaction progress.

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Most popular questions from this chapter

A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and \(27.0^{\circ} \mathrm{C}\). (a) Calculate the density of the gas in grams per liter. (b) What is the molar mass of the gas?

A sample of nitrogen gas kept in a container of volume \(2.3 \mathrm{~L}\) and at a temperature of \(32^{\circ} \mathrm{C}\) exerts a pressure of 4.7 atm. Calculate the number of moles of gas present.

State the following gas laws in words and also in the form of an equation: Boyle's law, Charles' law, Avogadro's law. In each case, indicate the conditions under which the law is applicable, and give the units for each quantity in the equation.

Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) burns in air: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ Balance the equation and determine the volume of air in liters at \(35.0^{\circ} \mathrm{C}\) and \(790 \mathrm{mmHg}\) required to burn \(227 \mathrm{~g}\) of ethanol. Assume that air is 21.0 percent \(\mathrm{O}_{2}\) by volume.

Acidic oxides such as carbon dioxide react with basic oxides like calcium oxide (CaO) and barium oxide \((\mathrm{BaO})\) to form salts (metal carbonates). (a) Write equations representing these two reactions. (b) A student placed a mixture of \(\mathrm{BaO}\) and \(\mathrm{CaO}\) of combined mass \(4.88 \mathrm{~g}\) in a 1.46 - \(\mathrm{L}\) flask containing carbon dioxide gas at \(35^{\circ} \mathrm{C}\) and \(746 \mathrm{mmHg}\). After the reactions were complete, she found that the \(\mathrm{CO}_{2}\) pressure had dropped to \(252 \mathrm{mmHg}\). Calculate the percent composition by mass of the mixture. Assume volumes of the solids are negligible.

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