World of Chemistry
World of Chemistry
7th Edition
ISBN: 9780618562763
Author: Steven S. Zumdahl
Publisher: Houghton Mifflin College Div
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Chapter 7, Problem 35A

(a)

Interpretation Introduction

Interpretation:

The balancing of given chemical equation has to be done.

  SiI4(s)+Mg(s)Si(s)+MgI2(s)

Concept Introduction :

Balancing of chemical equation is important to obey the law of conservation of mass.

(a)

Expert Solution
Check Mark

Answer to Problem 35A

The balanced chemical equation for given reaction is shown below:

  SiI4(s)+2Mg(s)Si(s)+2MgI2(s)

Explanation of Solution

The given unbalanced chemical equation is shown below:

  SiI4(s)+Mg(s)Si(s)+MgI2(s)

A coefficient 2 is used and MgI2 at right hand side is multiplied with it.

  SiI4(s)+Mg(s)Si(s)+2MgI2(s)

A coefficient 2 is used and Mg at left hand side is multiplied with it.

  SiI4(s)+2Mg(s)Si(s)+2MgI2(s)

The number of silicon atom is one and iodine atoms are four on both the sides of the equation. The number of atoms of magnesium is two on both the sides. Hence, the above equation is a balanced chemical equation.

(b)

Interpretation Introduction

Interpretation:

The balancing of given chemical equation has to be done. MnO2(s)+2Mg(s)Mn(s)+2MgO(s)

Concept Introduction :

Balancing of chemical equation is important to obey the law of conservation of mass.

(b)

Expert Solution
Check Mark

Answer to Problem 35A

The balanced chemical equation for given reaction is shown below:

  MnO2(s)+2Mg(s)Mn(s)+2MgO(s)

Explanation of Solution

The given chemical equation is shown below:

  MnO2(s)+Mg(s)Mn(s)+MgO(s)

A coefficient 2 is used and MgO at right hand side is multiplied with it.

  MnO2(s)+Mg(s)Mn(s)+2MgO(s)

A coefficient 2 is used and Mg at left hand side is multiplied with it.

  MnO2(s)+2Mg(s)Mn(s)+2MgO(s)

The number of magnesium and oxygen atoms is two on both the sides of the equation. The number of manganese atoms is one on both the sides of equation. Hence, the above equation is a balanced chemical equation.

(c)

Interpretation Introduction

Interpretation:

The balancing of given chemical equation has to be done.

  Ba(s)+S8(aq)BaS(s)

Concept Introduction :

Balancing of chemical equation is important to obey the law of conservation of mass.

(c)

Expert Solution
Check Mark

Answer to Problem 35A

The balanced chemical equation for given reaction is shown below:

  8Ba(s)+S8(aq)8BaS(s)

Explanation of Solution

The given chemical equation is shown below:

  Ba(s)+S8(aq)BaS(s)

A coefficient 8 is used and BaS at right hand side is multiplied with it.

  Ba(s)+S8(aq)8BaS(s)

A coefficient 8 is used and Ba at left hand side is multiplied with it.

  8Ba(s)+S8(aq)8BaS(s)

The number of barium and sulphur atoms is eight on both the sides of the equation. Hence, above equation is balanced.

(d)

Interpretation Introduction

Interpretation:

The balancing of given chemical equation has to be done.

  NH3(g)+Cl2(g)NH4Cl(s)+NCl3(g)

Concept Introduction :

Balancing of chemical equation is important to obey the law of conservation of mass.

(d)

Expert Solution
Check Mark

Answer to Problem 35A

The balanced chemical equation for given reaction is shown below:

  4NH3(g)+3Cl2(g)3NH4Cl(s)+NCl3(g)

Explanation of Solution

The given chemical equation is shown below:

  NH3(g)+Cl2(g)NH4Cl(s)+NCl3(g)

The number of nitrogen atom is one at left hand side and two at right hand side of the equation. The number of chlorine atom is two at left hand side and four at right hand side of the equation. The number of hydrogen atom is three at the left hand side and four at the right hand side of the equation. Hence, above equation is unbalanced.

In above equation NH4Cl and Cl2 is multiplied with a factor of 3 and NH3 with a factor of 4.

  4NH3(g)+3Cl2(g)3NH4Cl(s)+NCl3(g)

The number of nitrogen atom is four on both the side of the equation. The number of chlorine atom is sixon both the side of the equation. The number of hydrogen atom is twelveon both the side of the equation. Hence, above equation is balanced.

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