Glencoe Chemistry: Matter and Change, Student Edition
Glencoe Chemistry: Matter and Change, Student Edition
1st Edition
ISBN: 9780076774609
Author: McGraw-Hill Education
Publisher: MCGRAW-HILL HIGHER EDUCATION
Question
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Chapter 11.1, Problem 1PP

(a)

Interpretation Introduction

Interpretation:

A balanced chemical equation is given. It has to be interpreted in terms of particles, moles and mass and obeying of law of conservation of mass in it has to be shown.

N2(g)+3H2(g)2NH3(g)

Concept introduction:

According to law of conservation of mass, it is not possible to form the mass or to destroy the mass. The mass always remains conserved.

The number of moles is calculated as shown below:

Number of moles=Given massMolar mass (1)

(a)

Expert Solution
Check Mark

Answer to Problem 1PP

The given equation shows that one molecule of nitrogen gas reacts with three molecules of hydrogen gas to form two molecules of ammonia gas. In terms of number of moles it is seen that one mole of nitrogen gas reacts with three moles of hydrogen gas to form two moles of ammonia gas. Here, 28.0g nitrogen reacts with 6.0g of hydrogen to form 34.0g of ammonia. The law of conservation of mass is obeyed by the given chemical equation.

Explanation of Solution

The given chemical equation is shown below:

N2(g)+3H2(g)2NH3(g)

The above equation shows that one molecule of nitrogen gas reacts with three molecules of hydrogen gas to form two molecules of ammonia gas. In terms of number of moles, it is seen that one mole of nitrogen gas reacts with three moles of hydrogen gas to form two moles of ammonia gas.

The molar mass of nitrogen is 28.0g/mol. The mass of nitrogen is calculated as shown below:

1mol=Given mass28.0gmolGiven mass=1mol×28.0gmol=28.0g

The molar mass of hydrogen is 2.0g/mol. The mass of hydrogen is calculated as shown below:

3mol=Given mass2.0gmolGiven mass=3mol×2.0gmol=6.0g

The molar mass of ammonia is 17.0g/mol. The mass of ammonia is calculated as shown below:

2mol=Given mass17.0gmolGiven mass=2mol×17.0gmol=34.0g

Thus, 28.0g nitrogen reacts with 6.0g of hydrogen to form 34.0g of ammonia.

Total mass of reactants and products is given as shown below:

Total mass of reactants=Mass of N2+Mass of H2=28.0g+6.0g=34.0g

Total mass of products=Mass of NH3=34.0g

Here, the mass of reactants is equal to mass of the products. Thus, law of conservation of mass is obeyed.

(b)

Interpretation Introduction

Interpretation:

A balanced chemical equation is given. It has to be interpreted in terms of particles, moles and mass and obeying of law of conservation of mass in it has to be shown.

HCl(aq)+KOH(aq)KCl(aq)+H2O(l)

Concept introduction:

According to law of conservation of mass, it is not possible to form the mass or to destroy the mass. The mass always remains conserved.

The number of moles is calculated as shown below:

Number of moles=Given massMolar mass (1)

(b)

Expert Solution
Check Mark

Answer to Problem 1PP

In the given equation shows that one molecule of aqueous hydrogen chloride reacts with one molecule of potassium hydroxide to form one molecule of potassium chloride and one molecule of water. In terms of number of moles, it is seen that one mole of aqueous hydrogen chloride reacts with one mole of potassium hydroxide to form one mole of potassium chloride and one mole of water. Here, 36.46g aqueous hydrogen chloride reacts with 56.106g of potassium hydroxide to form 18.0g of water and 74.55g of potassium chloride. The law of conservation of mass is obeyed by the given chemical equation.

Explanation of Solution

The given chemical equation is shown below:

HCl(aq)+KOH(aq)KCl(aq)+H2O(l)

The above equation shows that one molecule of aqueous hydrogen chloride reacts with one molecule of potassium hydroxide to form one molecule of potassium chloride and one molecule of water. In terms of number of moles, it is seen that one mole of aqueous hydrogen chloride reacts with one mole of potassium hydroxide to form one mole of potassium chloride and one mole of water.

The molar mass of hydrogen chloride is 36.46g/mol. The mass of hydrogen chloride is calculated as shown below:

1mol=Given mass36.46gmolGiven mass=1mol×36.46gmol=36.46g

The molar mass of potassium hydroxide is 56.106g/mol. The mass of potassium hydroxide is calculated as shown below:

1mol=Given mass56.106gmolGiven mass=1mol×56.106gmol=56.106g

The molar mass of potassium chloride is 74.55g/mol. The mass of potassium chloride is calculated as shown below:

1mol=Given mass74.55gmolGiven mass=1mol×74.55gmol=74.55g

The molar mass of water is 18.0g/mol. The mass of water is calculated as shown below:

1mol=Given mass18.0gmolGiven mass=1mol×18.0gmol=18.0g

Thus, 36.46g aqueous hydrogen chloride reacts with 56.106g of potassium hydroxide to form 18.0g of water and 74.55g of potassium chloride.

Total mass of reactants and products is given as shown below:

Total mass of reactants=Mass of HCl+Mass of KOH=36.46g+56.106g=92.566g

Total mass of products=Mass of KCl+Mass of H2O=74.55g+18.0g=92.55g

Here, the mass of reactants is equal to mass of the products. Thus, law of conservation of mass is obeyed.

(c)

Interpretation Introduction

Interpretation:

A balanced chemical equation is given. It has to be interpreted in terms of particles, moles and mass and obeying of law of conservation of mass in it has to be shown.

2 Mg(s)+O2(g)2MgO(s)

Concept introduction:

According to law of conservation of mass it is not possible to form the mass or to destroy the mass. The mass always remains conserved.

The number of moles is calculated as shown below:

Number of moles=Given massMolar mass (1)

(c)

Expert Solution
Check Mark

Answer to Problem 1PP

The given equation shows that two molecules of magnesium react with one molecule of oxygen to form two molecule of magnesium oxide. In terms of number of moles, it is seen that two moles of magnesium reacts with one mole of oxygen to form one mole of magnesium oxide. Here, 48.61g magnesium reacts with 32.0g of oxygen to form 80.6g of magnesium oxide. The law of conservation of mass is obeyed by the given chemical equation.

Explanation of Solution

The given chemical equation is shown below:

2 Mg(s)+O2(g)2MgO(s)

The above equation shows that two molecules of magnesium react with one molecule of oxygen to form two molecule of magnesium oxide. In terms of number of moles it is seen that two moles of magnesium reacts with one mole of oxygen to form one mole of magnesium oxide.

The molar mass of magnesium is 24.305g/mol. The mass of magnesium is calculated as shown below:

2mol=Given mass24.305gmolGiven mass=2mol×24.305gmol=48.61g

The molar mass of oxygen gas is 32.0g/mol. The mass of oxygen is calculated as shown below:

2mol=Given mass32.0gmolGiven mass=1mol×32.0gmol=32.0g

The molar mass of magnesium oxide is 40.30g/mol. The mass of magnesium oxide is calculated as shown below:

2mol=Given mass40.30gmolGiven mass=2mol×40.30gmol=80.6g

Thus, 48.61g magnesium reacts with 32.0g of oxygen to form 80.6g of magnesium oxide.

Total mass of reactants and products is given as shown below:

Total mass of reactants=Mass of Mg+Mass of O2=48.61g+32.0g=80.61g

Total mass of products=Mass of MgO=80.6g

Here, the mass of reactants is equal to mass of the products. Thus, law of conservation of mass is obeyed.

Chapter 11 Solutions

Glencoe Chemistry: Matter and Change, Student Edition

Ch. 11.2 - Prob. 11PPCh. 11.2 - Prob. 12PPCh. 11.2 - Prob. 13PPCh. 11.2 - Prob. 14PPCh. 11.2 - Prob. 15PPCh. 11.2 - Prob. 16PPCh. 11.2 - Prob. 17SSCCh. 11.2 - Prob. 18SSCCh. 11.2 - Prob. 19SSCCh. 11.2 - Prob. 20SSCCh. 11.2 - Prob. 21SSCCh. 11.2 - Prob. 22SSCCh. 11.3 - Prob. 23PPCh. 11.3 - Prob. 24PPCh. 11.3 - Prob. 25SSCCh. 11.3 - Prob. 26SSCCh. 11.3 - Prob. 27SSCCh. 11.4 - Prob. 28PPCh. 11.4 - Prob. 29PPCh. 11.4 - Prob. 30PPCh. 11.4 - Prob. 31SSCCh. 11.4 - Prob. 32SSCCh. 11.4 - Prob. 33SSCCh. 11.4 - Prob. 34SSCCh. 11.4 - Prob. 35SSCCh. 11 - Prob. 36ACh. 11 - Prob. 37ACh. 11 - Prob. 38ACh. 11 - Prob. 39ACh. 11 - Prob. 40ACh. 11 - Prob. 41ACh. 11 - Prob. 42ACh. 11 - Prob. 43ACh. 11 - Interpret the following equation in terms of...Ch. 11 - Smelting When tin(IV) oxide is heated with carbon...Ch. 11 - When solid copper is added to nitric acid, copper...Ch. 11 - When hydrochloric acid solution reacts with lead...Ch. 11 - When aluminum is mixed with iron(lll) oxide, iron...Ch. 11 - Solid silicon dioxide, often called silica, reacts...Ch. 11 - Prob. 50ACh. 11 - Prob. 51ACh. 11 - Prob. 52ACh. 11 - Antacids Magnesium hydroxide is an ingredient in...Ch. 11 - Prob. 54ACh. 11 - Prob. 55ACh. 11 - Prob. 56ACh. 11 - Prob. 57ACh. 11 - Prob. 58ACh. 11 - Prob. 59ACh. 11 - Ethanol (C2H5OH) , also known as grain alcohol,...Ch. 11 - Welding If 5.50 mol of calcium carbide (CaC2)...Ch. 11 - Prob. 62ACh. 11 - Prob. 63ACh. 11 - Prob. 64ACh. 11 - Prob. 65ACh. 11 - Prob. 66ACh. 11 - Prob. 67ACh. 11 - Prob. 68ACh. 11 - Prob. 69ACh. 11 - Prob. 70ACh. 11 - Prob. 71ACh. 11 - Prob. 72ACh. 11 - Prob. 73ACh. 11 - Prob. 74ACh. 11 - Prob. 75ACh. 11 - Prob. 76ACh. 11 - Prob. 77ACh. 11 - Prob. 78ACh. 11 - Prob. 79ACh. 11 - Prob. 80ACh. 11 - Prob. 81ACh. 11 - Prob. 82ACh. 11 - Prob. 83ACh. 11 - Prob. 84ACh. 11 - Prob. 85ACh. 11 - Prob. 86ACh. 11 - Prob. 87ACh. 11 - Prob. 88ACh. 11 - Prob. 89ACh. 11 - Prob. 90ACh. 11 - Lead(ll) oxide is obtained by roasting galena,...Ch. 11 - Prob. 92ACh. 11 - Prob. 93ACh. 11 - Prob. 94ACh. 11 - Prob. 95ACh. 11 - Prob. 96ACh. 11 - Prob. 97ACh. 11 - Ammonium sulfide reacts With copper(ll) nitrate in...Ch. 11 - Fertilizer The compound calcium cyanamide (CaNCN)...Ch. 11 - When copper(ll) oxide is heated in the presence Of...Ch. 11 - Air Pollution Nitrogen monoxide, which is present...Ch. 11 - Electrolysis Determine the theoretical and percent...Ch. 11 - Iron reacts with oxygen as Shown....Ch. 11 - Analyze and Conclude In an experiment, you obtain...Ch. 11 - Observe and Infer Determine whether each reaction...Ch. 11 - Design an Experiment Design an experiment that can...Ch. 11 - Apply When a campfire begins to die down and...Ch. 11 - Apply Students conducted a lab to investigate...Ch. 11 - When 9.59 g of a certain vanadium oxide is heated...Ch. 11 - Prob. 110ACh. 11 - Prob. 111ACh. 11 - Prob. 112ACh. 11 - Prob. 113ACh. 11 - Prob. 114ACh. 11 - Prob. 115ACh. 11 - Prob. 116ACh. 11 - Prob. 117ACh. 11 - Prob. 118ACh. 11 - Prob. 119ACh. 11 - Prob. 120ACh. 11 - Prob. 1STPCh. 11 - Prob. 2STPCh. 11 - Prob. 3STPCh. 11 - Prob. 4STPCh. 11 - Prob. 5STPCh. 11 - Prob. 6STPCh. 11 - Prob. 7STPCh. 11 - Prob. 8STPCh. 11 - Prob. 9STPCh. 11 - Prob. 10STPCh. 11 - Prob. 11STPCh. 11 - Prob. 12STPCh. 11 - Prob. 13STPCh. 11 - Prob. 14STPCh. 11 - Prob. 15STPCh. 11 - Prob. 16STPCh. 11 - Prob. 17STP
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