/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 103 A mixture of \(\mathrm{N}_{2}(g)... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A mixture of \(\mathrm{N}_{2}(g)\) and \(\mathrm{H}_{2}(g)\) reacts in a closed container to form ammonia, \(\mathrm{NH}_{3}(g)\). The reaction ceases before either reactant has been totally consumed. At this stage \(3.0 \mathrm{~mol} \mathrm{} \mathrm{N}_{2}, 3.0 \mathrm{~mol} \mathrm{H}\), and \(3.0 \mathrm{~mol} \mathrm{} \mathrm{NH}_{3}\) are present. How many moles of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) were present originally?

Short Answer

Expert verified
Initially, there were 4.5 moles of \(N_{2}\) and 7.5 moles of \(H_{2}\) present.

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation for the formation of ammonia from nitrogen gas and hydrogen gas is: \[N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}(g)\]
02

Determine the extent of reaction

At the end of the reaction, we have 3.0 moles of NH₃. According to the balanced chemical equation, every 2 moles of NH₃ produced require 1 mole of N₂ and 3 moles of H₂. Therefore, to produce 3.0 moles of NH₃, the reacted moles of N₂ and H₂ can be determined by: Reacted moles of N₂ = \( \frac{3.0 \, moles \, NH_{3}}{2 \,moles \, NH_{3}}\) x 1 mole N₂ = 1.5 moles N₂ Reacted moles of H₂ = \( \frac{3.0 \, moles \, NH_{3}}{2 \,moles \, NH_{3}}\) x 3 moles H₂ = 4.5 moles H₂
03

Calculate initial moles of Nâ‚‚ and Hâ‚‚

Since we know the reacted moles of Nâ‚‚ and Hâ‚‚, we can find the initial moles by adding those to the moles present at the end of the reaction. Initial moles of Nâ‚‚ = Reacted moles of Nâ‚‚ + Moles of Nâ‚‚ present at the end Initial moles of Nâ‚‚ = 1.5 moles + 3.0 moles Initial moles of Nâ‚‚ = 4.5 moles Initial moles of Hâ‚‚ = Reacted moles of Hâ‚‚ + Moles of Hâ‚‚ present at the end Initial moles of Hâ‚‚ = 4.5 moles + 3.0 moles Initial moles of Hâ‚‚ = 7.5 moles Hence, initially, there were 4.5 moles of Nâ‚‚ and 7.5 moles of Hâ‚‚.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Balanced Chemical Equation
Writing a balanced chemical equation is the first crucial step in solving any stoichiometry problem. It shows you the relationship between reactants and products, telling you exactly how much of each substance participates in the reaction. In this problem, the balanced chemical equation for ammonia synthesis is:\[N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}(g)\]This equation shows that one mole of nitrogen (\(N_{2}\)) reacts with three moles of hydrogen (\(H_{2}\)) to produce two moles of ammonia (\(NH_{3}\)). Each component of the equation must be in this exact proportion to reflect the conservation of mass, ensuring that no atoms are lost or gained during the reaction.
Ammonia Synthesis
Ammonia synthesis is an essential chemical process, mainly done using the Haber process. The goal is to combine nitrogen (from the air) with hydrogen (from natural gas) to produce ammonia. This process is critical for creating fertilizers and has industrial importance. In our reaction, we start by mixing nitrogen and hydrogen in a container. They then react to form ammonia gas. However, it is important to follow the correct stoichiometric proportions as defined by the balanced chemical equation to ensure efficiency. This reaction is limited by the availability of reactants and stops when one of them is consumed. In real industrial applications, unreacted nitrogen and hydrogen can even be recycled to improve yield.
Reactants and Products
Reactants are the starting materials in a chemical reaction, while products are the substances formed by the reaction. In the equation for ammonia synthesis, the reactants are nitrogen (\(N_{2}\)) and hydrogen (\(H_{2}\)), and the product is ammonia (\(NH_{3}\)).The transformation from reactants to products follows a fixed ratio (as shown in the balanced equation). Here:
  • 1 mole of nitrogen reacts with 3 moles of hydrogen.
  • This produces 2 moles of ammonia.
Understanding which substances are reactants and which are products is key to correctly applying stoichiometry in chemical calculations.
Mole Concept
The mole is a fundamental unit in chemistry used to express amounts of a chemical substance. It provides a bridge between the atomic scale and the macroscopic world. Using the mole concept, chemists can count specific numbers of atoms or molecules by weighing them. In this exercise, the mole concept helps us determine how much of each reactant was initially present based on the amount of product formed. For every 2 moles of ammonia produced, 1 mole of nitrogen and 3 moles of hydrogen are used. This arena of calculation allows you to deduce how much nitrogen and hydrogen were originally in the system by:
  • Calculating how much reacted to form ammonia.
  • Adding the remaining unreacted amounts to find starting totals.
Thus, the mole concept simplifies the balancing of equations and the quantitative interpretation of chemical reactions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Give the empirical formula of each of the following com\(\mathrm{H}\), and \(0.0065 \mathrm{~mol} \mathrm{O}\); (b) \(11.66 \mathrm{~g}\) iron and \(5.01 \mathrm{~g}\) oxygen; (c) \(40.0 \% \mathrm{C}, 6.7 \% \mathrm{H}\), and \(53.3 \% \mathrm{O}\) by mass.

When hydrogen sulfide gas is bubbled into a solution of sodium hydroxide, the reaction forms sodium sulfide and water. How many grams of sodium sulfide are formed if \(1.25 \mathrm{~g}\) of hydrogen sulfide is bubbled into a solution containing \(2.00 \mathrm{~g}\) of sodium hydroxide, assuming that the sodium sulfide is made in \(92.0 \%\) yield?

The complete combustion of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), a component of gasoline, proceeds as follows: $$ 2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \longrightarrow 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) How many moles of \(\mathrm{O}_{2}\) are needed to burn \(1.50 \mathrm{~mol}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}\) ? (b) How many grams of \(\mathrm{O}_{2}\) are needed to burn \(10.0 \mathrm{~g}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}\) ? (c) Octane has a density of \(0.692 \mathrm{~g} / \mathrm{mL}\) at \(20^{\circ} \mathrm{C}\). How many grams of \(\mathrm{O}_{2}\) are required to burn \(15.0 \mathrm{gal}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}\) (the capacity of an average fuel tank)? (d) How many grams of \(\mathrm{CO}_{2}\) are produced when \(15.0 \mathrm{gal}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}\) are combusted?

Determine the empirical and molecular formulas of each of the following substances: (a) Styrene, a compound substance used to make Styrofoam cups and insulation, contains \(92.3 \% \mathrm{C}\) and \(7.7 \% \mathrm{H}\) by mass and has a molar mass of \(104 \mathrm{~g} / \mathrm{mol}\). (b) Caffeine, a stimulant found in coffee, contains \(49.5 \% \mathrm{C}\), \(5.15 \% \mathrm{H}, 28.9 \% \mathrm{~N}\), and \(16.5 \% \mathrm{O}\) by mass and has a molar mass of \(195 \mathrm{~g} / \mathrm{mol}\). (c) Monosodium glutamate (MSG), a flavor enhancer in certain foods, contains \(35.51 \% \mathrm{C}, 4.77 \% \mathrm{H}, 37.85 \% \mathrm{O}\), \(8.29 \% \mathrm{~N}\), and \(13.60 \% \mathrm{Na}\), and has a molar mass of \(169 \mathrm{~g} / \mathrm{mol}\).

Balance the following equations: (a) \(\mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)\) (b) \(\mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HNO}_{3}(a q)\) (c) \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(l)+\operatorname{HCl}(g)\) (d) \(\mathrm{Zn}(\mathrm{OH})_{2}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.