Lab 5

raymondbrousseauchem90

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Hi ray,

 

please see to attached file. This was the one I submitted. Thanks

  • Journal
  • FOCUS QUESTION: How and why are chemical equations balanced?

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    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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