Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 84

Ammonia and sulfuric acid react to form ammonium sulfate. (a) Write an equation for the reaction. (b) Determine the starting mass (in grams) of each reactant if \(20.3 \mathrm{~g}\) of ammonium sulfate is produced and \(5.89 \mathrm{~g}\) of sulfuric acid remains unreacted.

Short Answer

Expert verified
a) The equation for the reaction is \(2NH_3 + H_2SO_4 → (NH_4)_2SO_4\). b) Calculate the initial mass of sulfuric acid (\(H_2SO_4\)) first and then use this value to find the initial mass of ammonia (\(NH_3\)).

Step by step solution

01

Writing the Reaction Equation

The reaction equation for ammonia (\(NH_3\)) and sulfuric acid (\(H_2SO_4\)) forming ammonium sulfate (\((NH_4)_2SO_4\)) is \(2NH_3 + H_2SO_4 → (NH_4)_2SO_4\). It is crucial to ensure the equation is balanced, which in this case it is.
02

Calculation of Molar Masses

To manipulate exact amounts of reactants and products, convert the given masses into moles using their respective molar masses. The molar mass of \(NH_3\) is 17.03 g/mol, of \(H_2SO_4\) is 98.08 g/mol, and of \((NH_4)_2SO_4\) is 132.14 g/mol.
03

Calculation of initial Mass of Sulfuric Acid

From the stoichiometry of the reaction, we know that 1 mole of \(H_2SO_4\) yields 1 mole of \((NH_4)_2SO_4\). Therefore the initial moles of \(H_2SO_4\) is the sum of the moles of \((NH_4)_2SO_4\) produced and the moles of \(H_2SO_4\) remained unreacted. That is \( \frac{20.3g}{132.14g/mol} + \frac{5.89g}{98.08g/mol}\) moles. The initial mass of \(H_2SO_4\) is then this mole number times the molar mass of \(H_2SO_4\).
04

Calculation of initial Mass of Ammonia

According to the stoichiometry of the reaction, 2 moles of \(NH_3\) react with 1 mole of \(H_2SO_4\) to form 1 mole of \((NH_4)_2SO_4\). Hence the initial moles of \(NH_3\) is twice the initial moles of \(H_2SO_4\), calculated in step 3. The initial mass of \(NH_3\) is then this mole number times the molar mass of \(NH_3\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

A sample of a compound of \(\mathrm{Cl}\) and \(\mathrm{O}\) reacts with an excess of \(\mathrm{H}_{2}\) to give \(0.233 \mathrm{~g}\) of \(\mathrm{HCl}\) and \(0.403 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) Determine the empirical formula of the \(\mathrm{com}-\) pound.

limestone \(\left(\mathrm{CaCO}_{3}\right)\) is decomposed by heating to quicklime \((\mathrm{CaO})\) and carbon dioxide. Calculate how many grams of quicklime can be produced from \(1.0 \mathrm{~kg}\) of limestone.

What is wrong or ambiguous with each of the statements here? (a) \(\mathrm{NH}_{4} \mathrm{NO}_{2}\) is the limiting reactant in the reaction$$\mathrm{NH}_{4} \mathrm{NO}_{2}(s) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ (b) The limiting reactants for the reaction shown here are \(\mathrm{NH}_{3}\) and \(\mathrm{NaCl}\). $$\begin{aligned}\mathrm{NH}_{3}(a q)+\mathrm{NaCl}(a q) &+\mathrm{H}_{2} \mathrm{CO}_{3}(a q) & \longrightarrow \\ & \mathrm{NaHCO}_{3}(a q)+\mathrm{NH}_{4} \mathrm{Cl}(a q) \end{aligned}$$

The empirical formula of a compound is CH. If the molar mass of this compound is about \(78 \mathrm{~g},\) what is its molecular formula?

Ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right),\) an important industrial organic chemical, can be prepared by heating hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\) at \(800^{\circ} \mathrm{C}:$$$\mathrm{C}_{6} \mathrm{H}_{14} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}+\text { other products }$$ If the yield of ethylene production is 42.5 percent, what mass of hexane must be reacted to produce \)481 \mathrm{~g}$ of ethylene?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation
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.

Sign-up for free