/*! 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 98 Give the balanced equation for e... [FREE SOLUTION] | 91影视

91影视

Give the balanced equation for each of the following. a. The combustion of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) forms carbon dioxide and water vapor. A combustion reaction refers to a reaction of a substance with oxygen gas. b. Aqueous solutions of lead(Il) nitrate and sodium phosphate are mixed, resulting in the precipitate formation of lead(II) phosphate with aqueous sodium nitrate as the other product. c. Solid zinc reacts with aqueous HCl to form aqueous zinc chloride and hydrogen gas. d. Aqueous strontium hydroxide reacts with aqueous hydrobromic acid to produce water and aqueous strontium bromide.

Short Answer

Expert verified
a. \[ C_{2}H_{5}OH + \frac{3}{2}O_{2} \rightarrow 2CO_{2} + 3H_{2}O \] b. \[ 3Pb(NO_{3})_{2} + 2Na_{3}PO_{4} \rightarrow Pb_{3}(PO_{4})_{2} + 6NaNO_{3} \] c. \[ Zn + 2HCl \rightarrow ZnCl_{2} + H_{2} \] d. \[ Sr(OH)_{2} + 2HBr \rightarrow 2H_{2}O + SrBr_{2} \]

Step by step solution

01

a. Combustion of Ethanol

1. Write the unbalanced equation: Ethanol (C2H5OH) reacts with oxygen (O2) to form carbon dioxide (CO2) and water (H2O). \[ C_{2}H_{5}OH + O_{2} \rightarrow CO_{2} + H_{2}O \] 2. Balance the equation: To balance this equation, we need to ensure that the number of each type of atom on the reactants side is equal to the number of each type of atom on the products side. \[ C_{2}H_{5}OH + \frac{3}{2}O_{2} \rightarrow 2CO_{2} + 3H_{2}O \] The balanced equation for the combustion of ethanol is: \[ C_{2}H_{5}OH + \frac{3}{2}O_{2} \rightarrow 2CO_{2} + 3H_{2}O \]
02

b. Precipitate formation between lead(II) nitrate and sodium phosphate

1. Write the unbalanced equation: Lead(II) nitrate (Pb(NO3)2) reacts with sodium phosphate (Na3PO4) to form lead(II) phosphate (Pb3(PO4)2) and sodium nitrate (NaNO3). \[ Pb(NO_{3})_{2} + Na_{3}PO_{4} \rightarrow Pb_{3}(PO_{4})_{2} + NaNO_{3} \] 2. Balance the equation: To balance the equation, ensure that the number of each type of atom on the reactants side is equal to the number of each type of atom on the products side. \[ 3Pb(NO_{3})_{2} + 2Na_{3}PO_{4} \rightarrow Pb_{3}(PO_{4})_{2} + 6NaNO_{3} \] The balanced equation for the precipitate formation between lead(II) nitrate and sodium phosphate is: \[ 3Pb(NO_{3})_{2} + 2Na_{3}PO_{4} \rightarrow Pb_{3}(PO_{4})_{2} + 6NaNO_{3} \]
03

c. Reaction of solid zinc with aqueous HCl

1. Write the unbalanced equation: Solid zinc (Zn) reacts with aqueous HCl (HCl) to form aqueous zinc chloride (ZnCl2) and hydrogen gas (H2). \[ Zn + HCl \rightarrow ZnCl_{2} + H_{2} \] 2. Balance the equation: To balance the equation, ensure that the number of each type of atom on the reactants side is equal to the number of each type of atom on the products side. \[ Zn + 2HCl \rightarrow ZnCl_{2} + H_{2} \] The balanced equation for the reaction of solid zinc with aqueous HCl is: \[ Zn + 2HCl \rightarrow ZnCl_{2} + H_{2} \]
04

d. Reaction of aqueous strontium hydroxide with aqueous hydrobromic acid

1. Write the unbalanced equation: Aqueous strontium hydroxide (Sr(OH)2) reacts with aqueous hydrobromic acid (HBr) to form water (H2O) and aqueous strontium bromide (SrBr2). \[ Sr(OH)_{2} + HBr \rightarrow H_{2}O + SrBr_{2} \] 2. Balance the equation: To balance the equation, ensure that the number of each type of atom on the reactants side is equal to the number of each type of atom on the products side. \[ Sr(OH)_{2} + 2HBr \rightarrow 2H_{2}O + SrBr_{2} \] The balanced equation for the reaction of aqueous strontium hydroxide with aqueous hydrobromic acid is: \[ Sr(OH)_{2} + 2HBr \rightarrow 2H_{2}O + SrBr_{2} \]

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 91影视!

Key Concepts

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

Combustion Reaction
Combustion reactions are a fascinating type of chemical reaction where a substance combines with oxygen to produce energy in the form of heat or light. One common example is the combustion of ethanol, (\( \text{C}_{2}\text{H}_{5}\text{OH} \) ), which forms carbon dioxide and water. When balancing a combustion reaction, it's crucial to ensure that all atoms are equally accounted for on both sides of the equation.

In the case of ethanol, the balanced equation \( \text{C}_{2}\text{H}_{5}\text{OH} + \frac{3}{2}\text{O}_{2} \rightarrow 2\text{CO}_{2} + 3\text{H}_{2}\text{O} \) illustrates that one molecule of ethanol burns with one and a half molecules of oxygen to produce two molecules of carbon dioxide and three molecules of water.

Key Points:
  • Combustion involves oxygen as a reactant.
  • It often releases energy in the form of heat and light.
  • Balancing is about equalizing the atoms on both sides.
Precipitate Formation
Precipitate formation occurs when two aqueous solutions react to form an insoluble solid, known as a precipitate. This is a hallmark of many double displacement reactions. A classic example involves mixing lead(II) nitrate with sodium phosphate, which results in the formation of lead(II) phosphate as a precipitate, and sodium nitrate remains dissolved.

The balanced equation is \( 3\text{Pb(NO}_{3}\text{)}_{2} + 2\text{Na}_{3}\text{PO}_{4} \rightarrow \text{Pb}_{3}(\text{PO}_{4})_{2} + 6\text{NaNO}_{3} \) , showcasing how the reactants exchange ions to form new products.

Key Points:
  • Precipitate formation often results from the reaction of two ionic compounds.
  • The solid precipitate is typically indicated by a change in the solution's appearance.
  • Balancing involves matching the number of atoms and charges on both sides.
Single Displacement Reaction
Single displacement reactions involve one element replacing another in a compound. This can be observed when solid zinc reacts with hydrochloric acid (HCl), replacing the hydrogen ions to form zinc chloride and releasing hydrogen gas.

In this reaction, the balanced equation is \( \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_{2} + \text{H}_{2} \) . Here, zinc effectively displaces hydrogen, changing the chemical makeup of the reactants and products.

Key Points:
  • One element displaces another in single displacement reactions.
  • It commonly involves metals and acids or salts.
  • Balancing ensures the conservation of mass and charge.
Acid-Base Reaction
Acid-base reactions are essential chemical reactions where an acid reacts with a base to produce water and a salt. This is beautifully illustrated when strontium hydroxide reacts with hydrobromic acid.

The balanced equation for this reaction is \( \text{Sr(OH)}_{2} + 2\text{HBr} \rightarrow 2\text{H}_{2}\text{O} + \text{SrBr}_{2} \) , indicating how the hydrogen ions combine with hydroxide ions to form water, while the remaining ions form the salt of strontium bromide.

Key Points:
  • Acid-base reactions generally produce water and a salt.
  • They typically involve the transfer of protons (H鈦 ions).
  • Balancing these reactions focuses on equalizing proton and hydroxide quantities.

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

Which of the following statements about chemical equations is(are) true? a. When balancing a chemical equation, you can never change the coefficient in front of any chemical formula. b. The coefficients in a balanced chemical equation refer to the number of grams of reactants and products. c. In a chemical equation, the reactants are on the right and the products are on the left. d. When balancing a chemical equation, you can never change the subscripts of any chemical formula. e. In chemical reactions, matter is neither created nor destroyed so a chemical equation must have the same number of atoms on both sides of the equation.

Consider the following unbalanced reaction: $$ \mathrm{P}_{4}(s)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{PF}_{3}(g) $$ What mass of \(\mathrm{F}_{2}\) is needed to produce \(120 . \mathrm{g}\) of \(\mathrm{PF}_{3}\) if the reaction has a 78.1\(\%\) yield?

A sample of a hydrocarbon (a compound consisting of only carbon and hydrogen contains \(2.59 \times 10^{23}\) atoms of hydrogen and is 17.3\(\%\) hydrogen by mass. If the molar mass of the hydrocarbon is between 55 and 65 g/mol, what amount (moles) of compound is present, and what is the mass of the sample?

Phosphorus can be prepared from calcium phosphate by the following reaction: $$ 2 \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+6 \mathrm{SiO}_{2}(s)+10 \mathrm{C}(s) \longrightarrow $$ $$ 6 \mathrm{CaSiO}_{3}(s)+\mathrm{P}_{4}(s)+10 \mathrm{CO}(g) $$ Phosphorite is a mineral that contains \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) plus other non-phosphorus-containing compounds. What is the maximum amount of \(\mathrm{P}_{4}\) that can be produced from 1.0 \(\mathrm{kg}\) of phosphorite if the phorphorite sample is 75\(\% \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) by mass? Assume an excess of the other reactants.

You may have noticed that water sometimes drips from the exhaust of a car as it is running. Is this evidence that there is at least a small amount of water originally present in the gasoline? Explain.

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.