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Ira Lisetskai [31]
2 years ago
9

If 4.88 grams of zn react with 5.03 grams of s8 to produce 6.02 grams of zns, what are the theoretical yield and percent yield o

f this reaction? be sure to show the work that you did to solve this problem. unbalanced equation: zn + s8 zns
Chemistry
1 answer:
nataly862011 [7]2 years ago
8 0
<span>Balance the equation
    8Zn + S8 -----> 8ZnS
    work out how much ZnS is possible if all of each reagent were to react. The reagent that produces the least product is the limiting reagent.
    moles Zn = mass / molar mass = 4.88 g / 65.38 g/mol = 0.07464 moles
  8 moles Zn reacts to form 8 moles ZnS
  Therefore moles ZnS possible = moles Zn = 0.07464 moles

    moles S8 = 5.03 g / 256.48 g/mol = 0.01961 mol
  1 mole S8 reacts to form 8 moles ZnS
  Therefore moles ZnS possible = 8 x moles S8 = 0.1569 mol

    The amount of Zn provided gives the least ZnS, therefore Zn is the limiting reagent.
    The theoretical yield is the maximum possible yield possible from the reagents provided. It is the amount of product that will form if all the limiting reagent reacts.
  So in this case 4.88 g Zn yields 0.07464 mol ZnS

    theoretical mass ZnS = molar mass x moles
  = 97.44 g/mol x 0.07464 mol
  = 7.27 g

    % yield = actual yield / theoretical yield x 100/1
  = 6.02 g / 7.27 g x 100/1
  = 82.8 %</span>
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For the following reactions, predict the products and write the balanced formula equation, complete ionic equation, and net ioni
stealth61 [152]

Answer:

.

Explanation:

To predict the products of these reactions we need to know the kind of reactions. All these reactions are double replacement reaction. In these kinds of reactions, the products will be the result of exchanging ions in the reactants. So, the first step is to identify the ions.  

For the reaction, we have Hg2(NO3)2 and CuSO4.  We have the ions Hg+1,  NO3-1,   Cu+2 and SO4-2  

The way to make this exchange is putting together positive in one species with the negative of the other species. Following that rule we have

Hg^{+1}  - - -  (SO_{4})^{-2}[/text]&#10;the oxidation number will tell you the subscript for each species in the compound. In this case, is Hg2(SO4)  [tex]Cu^{+2} - - -  (NO_{3})^{-1}  - - ->  Cu(NO_{3})_{2} [/text]  &#10;So, the products for this reaction will be&#10;  [tex]Hg_{2} (NO_{3})_{2}(aq) + CuSO_{4}(aq)  -->  Hg_{2}SO_{4} + Cu(NO_{3})_{2}[/text]&#10;&#10;After this, we proceed to balance the equation. For this, we check that we have the same number of each element on both sides of the equation. In this case, we can see that we have the same number, so the equation is balanced.  Finally, we check the rules of solubility to see if the species are soluble in water or not. In this case sulfates area always soluble except for mercury so Hg2(SO4) precipitates in the solution (pre). Nitrates are always soluble so Cu(NO3)2 is soluble (aq)  &#10;[tex] Hg_{2}(NO_{3})_{2}(aq) + CuSO_{4}(aq)  - -> Hg_{2}SO_{4} (pre) + Cu(NO_{3})_{2}(aq)

The complete ionic equation allows to show which of the reactants or products exist primarily as ions.  For this reaction this will be:

2Hg^{+1}(aq)  + 2(NO_{3})^{-1}(aq) + (SO_{4})^{-2}(aq)  + Cu^{+2}(aq)    -->  Hg_{2}SO_{4} (pre)+ Cu^{+2}(aq)    + (NO_{3})^{-1}(aq) [/text]&#10;&#10;To get net ionic equation we take away the ions that did not participate in the reactions. In other words the ones that are the same on both sides in the equation. In this case we see that [tex] Cu^{+2}(aq)   and  (NO_{3})^{-1}(aq) [/text] are the same on both sides so those ions are not include in the net ionic equation.  This is:&#10;[tex] 2Hg^{+1}(aq)  + (SO_{4})^{-2}(aq)  -->  Hg_{2}SO_{4} (pre) [/text]&#10;&#10;B [tex] Ni(NO_{3})_{2}(aq) + CaCl_{2}(aq)

ions (1) Ni^{+2}  and (NO_{3})^{-1}

ions (2) Ca^{+2} and Cl^{-1}

Exchanging  

Ni^{+2}  ---- Cl^{-1}  -->  NiCl_{2}  

Ca^{+2} ---  (NO_{3})^{-1}  -->  Ca(NO_{3})_{2}  

Products  

Ni(NO_{3})_{2}(aq) + CaCl_{2}(aq) -->  NiCl_{2}  + Ca(NO_{3})_{2}  

The equation is already balanced

Chlorides are always soluble except Ag+, TI+, Pb+2 and Hg2+2. NiCl2 is soluble (aq)

Nitrates are always soluble. Ca(NO3)2 is soluble (aq)  

Since both compounds are soluble, we can say that there is not reaction.

Complete ionic equation  

Ni^{+2}(aq) + 2(NO_{3})^{-1}  (aq) + Ca^{+2}(aq) + 2Cl^{-1}(aq) - - > Ni^{+2}(aq) + 2(NO_{3})^{-1}  (aq) + Ca^{+2}(aq) + 2Cl^{-1}(aq)

Net ionic equation:

The ions in both sides of the equation are the same so all of them are cancelled and we cannot get a net ionic equation this explains why there is no reaction in this case.  

C K_{2}CO_{3}(aq) + MgI_{2}(aq)

Ions(1) K^{+1}  and (CO_{3})^{-2}

Ions(2) Mg^{+2}  and l^{-1}

Exchanging  

K^{+1}  ---  l^{-1}  - - >  KI

Mg^{+2}  ---  (CO_{3})^{-2}  - - >  Ca(CO_{3})

Products  

K_{2}CO_{3}(aq) + MgI_{2}(aq) - ->   Kl + MgCO_{3}  

The equation is not balanced

Balance equation is  

K_{2}CO_{3}(aq) + MgI_{2}(aq) - ->  2Kl (aq) + MgCO_{3} (pre)  

iodides are always soluble except Ag+, TI+, Pb+2 and Hg2+2. KI is soluble (aq)

carbonates are always insoluble except group 1 cations. MgCO3 is insoluble (pre)

complete ionic equation  

2K^{+1}(aq)  + (CO_{3})^{-2}(aq)  + Mg^{+2}(aq)   + 2l^{-1}(aq)  - - > MgCO_{3} (pre) + 2K^{+1}(aq)  + 2l^{-1}(aq)  

Net ionic equation

(CO_{3})^{-2}(aq)  + Mg^{+2}(aq)  - - > MgCO_{3} (pre)  

D Na_{2}CrO_{4}(aq) + AlBr_{3}(aq)  

Ions(1) Na^{+1}  and (CrO_{4})^{-2}

Ions(2) Al^{+3} and Br^{-1}

Exchanging  

Na^{+1}  ---- Br^{-1} - ->  NaBr  

Al^{+3} ---  (CrO_{4})^{-2} - ->  Al_{2}(CrO_{4})_{3}

Products  

Na_{2}CrO_{4}(aq) + AlBr_{3}(aq) - ->  NaBr  + Al_{2}(CrO_{4})_{3}

The equation is not balanced

Balance equation is  

3Na_{2}CrO_{4}(aq) + 2AlBr_{3}(aq) - -> 6NaBr  + Al_{2}(CrO_{4})_{3}

bromides are always soluble except Ag+, TI+, Pb+2 and Hg2+2. NaBr is soluble (aq)

chromates are always insoluble except group 1 cations. Al2(CrO4)3 is insoluble  (pre)

3Na_{2}CrO_{4}(aq) + 2AlBr_{3}(aq) - ->  6NaBr(aq) + Al_{2}(CrO_{4})_{3}(pre)

Complete ionic equation

6Na^{+1}(aq)  + 3(CrO_{4})^{-2}(aq) + 2Al^{+3}(aq) + 6Br^{-1}(aq) - -> Al_{2}(CrO_{4})_{3}(pre) +6Br^{-1}(aq) +  6Na^{+1}(aq)  

Net ionic equation

3(CrO_{4})^{-2}(aq) + 2Al^{+3}(aq) - -> Al_{2}(CrO_{4})_{3}(pre)  

6 0
3 years ago
In class we derived the Gibbs energy of mixing for a binary mixture of perfect gases. We also discussed that the same result is
GaryK [48]

Answer:

Attached below

Explanation:

Free energy of mixing = ΔGmix = Gf - Gi

attached below is the required derivation of the

<u>a) Molar Gibbs energy of mixing</u>

ΔGmix = Gf - Gi

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<u>b) molar excess Gibbs energy of mixing</u>

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fi = Fugacity

N°i = Chemical potential of gas when Fugacity = 1

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Which of these tasks falls under the responsibility of a forensic scientist?
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Answer:

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