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"; Lab 5-How And Why Are Chemical Equations Balanced - Chem Homework Help
Lab 5-How and why are chemical equations balanced

More Instructions

Journal

FOCUS QUESTION: How and why are chemical equations balanced?

TAB 1: Equations
Read the questions below. Then complete this Journal by interacting with the online Simulation.

Remember: the Journal does NOT check your answers. Review your text entries and make sure you’ve transferred data to the correct table
rows.

Chemical equations represent what happens during the course of chemical reactions. Reactants and
products, their quantities, and their physical states can be represented, as in the following equation:

Zn (s) + 2 HCl (aq)0020 2192→ ZnCl2 (aq) + H2 (g).

In words, solid zinc reacts with aqueous hydrochloric acid (hydrogen chloride) to yield aqueous zinc
chloride and hydrogen gas. The “2” in front of HCl is called a coefficient. Coefficients indicate the
simplest, whole-number ratios between the substances in the reaction. Their purpose is to ensure that
the equation is balanced and that matter is conserved. In a balanced equation, the number of atoms of
an element on the reactant side of the equation equals the number of atoms of that same element on
the product side.

In this section, you will learn how to count the number of atoms within a substance and to recognize balanced and unbalanced chemical
equations.

In the Simulation (to the right) select a reaction from the pull-down menu. Drag the molecules to the two-
pan balance to count the numbers of reactant and product atoms and to determine if the equation is
balanced.

1.1 Data Collection
For each chemical reaction:

• Pay attention to the Data panel as you drag each molecule to the balance. Observe how the atom counts change.
• Use the Transfer Data button ( ) to transfer the atom count from the Data panel to the table.
• Then, use the pull-down menus to indicate if each substance is balanced.

Reaction 1: CO + 2 H20020 2192→ CH3OH
elements: C O H

# of reactant atoms: 1 1 4

# of product atoms: 1 1 4

balanced? yes yes yes

Reaction 2: Fe2O3 + 3 Mg0020 2192→ MgO + 2 Fe
elements: Fe O Mg

# of reactant atoms: 2 3 3

# of product atoms: 2 1 1

balanced? yes no no

Reaction 3: 2 NH3 + O20020 2192→ 2 NO + 3 H2O
elements: N H O

# of reactant atoms: 2 6 2

# of product atoms: 2 6 5

balanced? yes yes no

Science 1194 SAS® Curriculum Pathways®

Chemical Equations: Journal

NAME: ray CLASS: chem90 DATE: 10/27/2013

Copyright © 2013, SAS Institute Inc., Cary, NC, USA, All Rights Reserved Page 1 of 6

Reaction 4: (NH4)2PtCl60020 2192→ 2 NH4Cl + Pt + 2 Cl2
elements: N H Pt Cl

# of reactant atoms: 2 8 1 6

# of product atoms: 2 8 1 6

balanced? yes yes yes yes

Reaction 5: 2 C2H6 + 7 O20020 2192→ 4 CO2 + 6 H2O
elements: C H O

# of reactant atoms: 4 12 14

# of product atoms: 4 12 14

balanced? yes yes yes

1.2 Name and state the law requiring chemical equations to be balanced (discussed in the Welcome video).

It is the law of conservation of mass. It states that the mass is neither created nor destroyed. Thus, the mass in the reactacnt
should be the same in the product.

1.3 For Reaction 2, putting a coefficient of 3 in front of MgO produces a balanced equation. Explain why the equation cannot be balanced by
changing MgO to Mg3O3.

Putting in front 3 MgO is technically not allowed. Besides the law of definite composition states that chemical compounds are
composed of fixed ratio of elements by mass. Thus there’s only MgO and not Mg3O3

1.4 Can Reaction 3 be balanced by putting a coefficient of 5/2 in front of O2? Why or why not?

It can be balanced but it is not customary to leave coeficients as fractions. We need to multiply coefficients in both sides by two as
per the law of multiple proportion which states that masses of one element which combine with a fixed mass of second element are in a
ratio of whole numbers.

1.5 In Reaction 4, 2 NH4Cl molecules are formed. In total, how many atoms of each element do these two molecules contain? Explain your
reasoning.

N = 2 H = 8 Cl = 2 .In doing so, we only need to multiply each coefficient with the subscript to obtain the number of atoms in each
element. So for N its 2×1 = 2. for H its 2×4=8 and for Cl its 2×1=2 .

1.6 In Reaction 5, the atom count for oxygen is 14. Explain how the product side of the equation represents 14 oxygen atoms.

In 4O2 the number of Oxygen atoms is 4×2 = 8
In 6H2O the number of oxygen atoms is 6×1 = 6
So 6+8 there are now 14 oxygen atoms in total

TAB 2: Balancing
In this section, you will develop your equation-balancing skills. The process of balancing a chemical equation could involve trial and error, but
having a procedure to follow makes the task simpler. To view a step-by-step process, click the “how-to” button.

In the Simulation (to the right) select a reaction from the pull-down menu. Balance the equation by
dragging substances to the reactant and product areas. The Simulation displays a Congratulations
message when the equation is correct. Pay close attention to the Data panel as you work.

Science 1194 SAS® Curriculum Pathways®

Copyright © 2013, SAS Institute Inc., Cary, NC, USA, All Rights Reserved Page 2 of 6

2.1 Data Collection
After balancing an equation, use the Transfer Data button ( ) to transfer both the equation and atom count from the Data panel to the table.

Reaction 1: 1 FeCl3 + 3 KOH0020 2192→ 3 KCl + 1 Fe(OH)3
elements: Fe Cl K O H

# of reactant atoms: 1 3 3 3 3

# of product atoms: 1 3 3 3 3

Reaction 2: 2 Na3PO4 + 3 Cu(NO3)20020 2192→ 1 Cu3(PO4)2 + 6 NaNO3
elements: Na P O Cu N

# of reactant atoms: 6 2 26 3 6

# of product atoms: 6 2 26 3 6

Reaction 3: 2 Y(OH)3 + 3 H2SO40020 2192→ 1 Y2(SO4)3 + 6 H2O
elements: Y O H S

# of reactant atoms: 2 18 12 3

# of product atoms: 2 18 12 3

Reaction 4: 4 PH3 + 8 O20020 2192→ 1 P4O10 + 6 H2O
elements: P H O

# of reactant atoms: 4 12 16

# of product atoms: 4 12 16

Reaction 5: 2 CH3OH + 3 O20020 2192→ 2 CO2 + 4 H2O
elements: C H O

# of reactant atoms: 2 8 8

# of product atoms: 2 8 8

Reaction 6: 1 C6H5CH3 + 9 O20020 2192→ 7 CO2 + 4 H2O
elements: C H O

# of reactant atoms: 7 8 18

# of product atoms: 7 8 18

TAB 3: Practice
In this section, you will continue to develop skills in generating balanced chemical equations.

In the Simulation (to the right) select a reaction from the pull-down menu. Balance the equation by
dragging substances to the reactant and product areas. The Simulation displays a Congratulations
message when the equation is correct.

Science 1194 SAS® Curriculum Pathways®

Copyright © 2013, SAS Institute Inc., Cary, NC, USA, All Rights Reserved Page 3 of 6

3.1 Data Collection
• Enter the atom counts in the table as you work to balance an equation.
• Use the Transfer Data button ( ) to transfer the balanced equation from the Data panel to the table.
• Make sure your final atom counts are based on the balanced equation.

Reaction 1: 2 Na + 2 H2O0020 2192→ 2 NaOH + 1 H2
elements: Na H O

# of reactant atoms: 2 4 2

# of product atoms: 2 4 2

Reaction 2: 1 Fe2O3 + 3 CO0020 2192→ 2 Fe + 3 CO2
elements: Fe O C

# of reactant atoms: 2 6 3

# of product atoms: 2 6 3

Reaction 3: 2 Fe(OH)3 + 3 H2S0020 2192→ 1 Fe2S3 + 6 H2O
elements: Fe O H S

# of reactant atoms: 2 6 12 3

# of product atoms: 2 6 12 3

Reaction 4: 1 P4S3 + 8 O20020 2192→ 1 P4O10 + 3 SO2
elements: P S O

# of reactant atoms: 4 3 16

# of product atoms: 4 3 16

Reaction 5: 4 KO2 + 2 CO20020 2192→ 2 K2CO3 + 3 O2
elements: K O C

# of reactant atoms: 4 12 2

# of product atoms: 4 12 2

Reaction 6: 1 Mg3N2 + 8 HCl0020 2192→ 3 MgCl2 + 2 NH4Cl
elements: Mg N H Cl

# of reactant atoms: 3 2 8 8

# of product atoms: 3 2 8 8

Reaction 7: 1 P4O10 + 12 HClO40020 2192→ 4 H3PO4 + 6 Cl2O7
elements: P O H Cl

# of reactant atoms: 4 58 12 12

# of product atoms: 4 58 12 12

Science 1194 SAS® Curriculum Pathways®

Copyright © 2013, SAS Institute Inc., Cary, NC, USA, All Rights Reserved Page 4 of 6

Analysis

TAB 4: Analysis
Refer to the Journal, as needed, to answer the following questions.

Balancing Equations
A.1 Using what you learned in the first three tabs, balance the following equations.

• Complete each table by entering the appropriate coefficients and atom counts.
• Coefficients of “1” do not need to be entered.

Reaction 1: 4 FeS2 + 11 O20020 2192→ 2 Fe2O3 + 8 SO2
elements: Fe S O

# of reactant atoms: 4 8 22

# of product atoms: 4 8 22

Reaction 2: 2 C7H6O2 + 15 O20020 2192→ 14 CO2 + 6 H2O
elements: C H O

# of reactant atoms: 14 12 34

# of product atoms: 14 12 34

Reaction 3: 4 C3H5NO + 19 O20020 2192→ 12 CO2 + 4 NO2 + 10 H2O
elements: C H N O

# of reactant atoms: 12 20 4 42

# of product atoms: 12 20 4 42

Reaction 4: C6H12O6 + 4 KClO30020 2192→ 6 CO2 + 6 H2O + 4 KCl
elements: C H O K Cl

# of reactant atoms: 6 12 18 4 4

# of product atoms: 6 12 18 4 4

Conclusion
A.2 Focus Question — How and why are chemical equations balanced? In answering this question, discuss:

• the reason equations must be balanced
• what you can change to balance an equation
• the process of equation-balancing

Finally, to demonstrate your understanding, balance the following equation (from the end of the Welcome video):

Cu(NO3)2 + Na3PO40020 2192→ NaNO3 + Cu3(PO4)2.

NOTE: you cannot type subscripts in the text box below. Cu(NO3)2, for example, would be entered as Cu(NO3)2.

Science 1194 SAS® Curriculum Pathways®

Chemical Equations: Analysis

NAME: ray CLASS: chem90 DATE: 10/27/2013

Copyright © 2013, SAS Institute Inc., Cary, NC, USA, All Rights Reserved Page 5 of 6

Answer to Focus Question:
Equations should be balanced in accordance with the law of conservation of mass which states that the mass is neither created nor
destroyed which means we have to balance equation such that the mass in the reactant is the same as the product. In balancing equation
it is the coefficients that we can change. To balance equation, we need to find the right coefficients in the reactants and products that will
yield the same number of atoms of each elements in each side.

Answer to balancing equation:

3 Cu(NO3)2 + 2Na3PO4 -> 6NaNO3 + Cu3(PO4)2

Science 1194 SAS® Curriculum Pathways®

Copyright © 2013, SAS Institute Inc., Cary, NC, USA, All Rights Reserved Page 6 of 6

  • Journal
    • TAB 1: Equations
    • TAB 2: Balancing
    • TAB 3: Practice
  • Analysis
    • TAB 4: Analysis
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