Is there a relationship between the mass of the reactants and the mass of the products if so what is the relationship?

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You have learned that when reactants are not present in stoichiometric quantities, the limiting reactant determines the maximum amount of product that can be formed from the reactants. The amount of product calculated in this way is the theoretical yieldThe maximum amount of product that can be formed from the reactants in a chemical reaction, which theoretically is the amount of product that would be obtained if the reaction occurred perfectly and the method of purifying the product were 100% efficient., the amount you would obtain if the reaction occurred perfectly and your method of purifying the product were 100% efficient.

In reality, you almost always obtain less product than is theoretically possible because of mechanical losses (such as spilling), separation procedures that are not 100% efficient, competing reactions that form undesired products, and reactions that simply do not go all the way to completion, thus resulting in a mixture of products and reactants. This last possibility is a common occurrence and is the subject of . So the actual yieldThe measured mass of products actually obtained from a reaction. The actual yield is nearly always less than the theoretical yield., the measured mass of products obtained from a reaction, is almost always less than the theoretical yield (often much less). The percent yieldThe ratio of the actual yield of a reaction to the theoretical yield multiplied by 100 to give a percentage. of a reaction is the ratio of the actual yield to the theoretical yield, multiplied by 100 to give a percentage:

The method used to calculate the percent yield of a reaction is illustrated in Example 13.

Procaine is a key component of Novocain, an injectable local anesthetic used in dental work and minor surgery. Procaine can be prepared in the presence of H2SO4 (indicated above the arrow) by the reaction

If we carried out this reaction using 10.0 g of p-aminobenzoic acid and 10.0 g of 2-diethylaminoethanol, and we isolated 15.7 g of procaine, what was the percent yield?

Given: masses of reactants and product

Asked for: percent yield

Strategy:

A Write the balanced chemical equation.

B Convert from mass of reactants and product to moles using molar masses and then use mole ratios to determine which is the limiting reactant. Based on the number of moles of the limiting reactant, use mole ratios to determine the theoretical yield.

C Calculate the percent yield by dividing the actual yield by the theoretical yield and multiplying by 100.

Solution:

A From the formulas given for the reactants and the products, we see that the chemical equation is balanced as written. According to the equation, 1 mol of each reactant combines to give 1 mol of product plus 1 mol of water.

B To determine which reactant is limiting, we need to know their molar masses, which are calculated from their structural formulas: p-aminobenzoic acid (C7H7NO2), 137.14 g/mol; 2-diethylaminoethanol (C6H15NO), 117.19 g/mol. Thus the reaction used the following numbers of moles of reactants:

The reaction requires a 1:1 mole ratio of the two reactants, so p-aminobenzoic acid is the limiting reactant. Based on the coefficients in the balanced chemical equation, 1 mol of p-aminobenzoic acid yields 1 mol of procaine. We can therefore obtain only a maximum of 0.0729 mol of procaine. To calculate the corresponding mass of procaine, we use its structural formula (C13H20N2O2) to calculate its molar mass, which is 236.31 g/mol.

C The actual yield was only 15.7 g of procaine, so the percent yield was

(If the product were pure and dry, this yield would indicate that we have very good lab technique!)

Exercise

Lead was one of the earliest metals to be isolated in pure form. It occurs as concentrated deposits of a distinctive ore called galena (PbS), which is easily converted to lead oxide (PbO) in 100% yield by roasting in air via the following reaction:

The resulting PbO is then converted to the pure metal by reaction with charcoal. Because lead has such a low melting point (327°C), it runs out of the ore-charcoal mixture as a liquid that is easily collected. The reaction for the conversion of lead oxide to pure lead is as follows:

If 93.3 kg of PbO is heated with excess charcoal and 77.3 kg of pure lead is obtained, what is the percent yield?

Answer: 89.2%

Percent yield can range from 0% to 100%. A 100% yield means that everything worked perfectly, and you obtained all the product that could have been produced. Anyone who has tried to do something as simple as fill a salt shaker or add oil to a car’s engine without spilling knows how unlikely a 100% yield is. At the other extreme, a yield of 0% means that no product was obtained. A percent yield of 80%–90% is usually considered good to excellent; a yield of 50% is only fair. In part because of the problems and costs of waste disposal, industrial production facilities face considerable pressures to optimize the yields of products and make them as close to 100% as possible.

1. Engineers use conservation of mass, called a “mass balance,” to determine the amount of product that can be obtained from a chemical reaction. Mass balance assumes that the total mass of reactants is equal to the total mass of products. Is this a chemically valid practice? Explain your answer.

2. Given the equation 2H2(g) + O2(g) → 2H2O(g), is it correct to say that 10 g of hydrogen will react with 10 g of oxygen to produce 20 g of water vapor?

3. What does it mean to say that a reaction is stoichiometric?

4. When sulfur is burned in air to produce sulfur dioxide, what is the limiting reactant? Explain your answer.

5. Is it possible for the percent yield to be greater than the theoretical yield? Justify your answer.

Please be sure you are familiar with the topics discussed in Essential Skills 2 () before proceeding to the Numerical Problems.

1. What is the formula mass of each species?

   1. ammonium chloride
   2. sodium cyanide
   3. magnesium hydroxide
   4. calcium phosphate
   5. lithium carbonate
   6. hydrogen sulfite ion

2. What is the molecular or formula mass of each compound?

   1. potassium permanganate
   2. sodium sulfate
   3. hydrogen cyanide
   4. potassium thiocyanate
   5. ammonium oxalate
   6. lithium acetate

3. How many moles are in each of the following?

   1. 10.76 g of Si
   2. 8.6 g of Pb
   3. 2.49 g of Mg
   4. 0.94 g of La
   5. 2.68 g of chlorine gas
   6. 0.089 g of As

4. How many moles are in each of the following?

   1. 8.6 g of CO2
   2. 2.7 g of CaO
   3. 0.89 g of KCl
   4. 4.3 g of SrBr2
   5. 2.5 g of NaOH
   6. 1.87 g of Ca(OH)2

5. Convert the following to moles and millimoles.

   1. 1.68 g of Ba(OH)2
   2. 0.792 g of H3PO4
   3. 3.21 g of K2S
   4. 0.8692 g of Cu(NO3)2
   5. 10.648 g of Ba3(PO4)2
   6. 5.79 g of (NH4)2SO4
   7. 1.32 g of Pb(C2H3O2)2
   8. 4.29 g of CaCl2·6H2O

6. Convert the following to moles and millimoles.

   1. 0.089 g of silver nitrate
   2. 1.62 g of aluminum chloride
   3. 2.37 g of calcium carbonate
   4. 1.004 g of iron(II) sulfide
   5. 2.12 g of dinitrogen pentoxide
   6. 2.68 g of lead(II) nitrate
   7. 3.02 g of ammonium phosphate
   8. 5.852 g of sulfuric acid
   9. 4.735 g of potassium dichromate

7. What is the mass of each substance in grams and milligrams?

   1. 5.68 mol of Ag
   2. 2.49 mol of Sn
   3. 0.0873 mol of Os
   4. 1.74 mol of Si
   5. 0.379 mol of H2
   6. 1.009 mol of Zr

8. What is the mass of each substance in grams and milligrams?

   1. 2.080 mol of CH3OH
   2. 0.288 mol of P4
   3. 3.89 mol of ZnCl2
   4. 1.800 mol of Fe(CO)5
   5. 0.798 mol of S8
   6. 4.01 mol of NaOH

9. What is the mass of each compound in kilograms?

   1. 6.38 mol of P4O10
   2. 2.26 mol of Ba(OH)2
   3. 4.35 mol of K3PO4
   4. 2.03 mol of Ni(ClO3)2
   5. 1.47 mol of NH4NO3
   6. 0.445 mol of Co(NO3)3

10. How many atoms are contained in each?

    1. 2.32 mol of Bi
    2. 0.066 mol of V
    3. 0.267 mol of Ru
    4. 4.87 mol of C
    5. 2.74 g of I2
    6. 1.96 g of Cs
    7. 7.78 g of O2

11. Convert each number of atoms to milligrams.

    1. 5.89 × 1022 Pt atoms
    2. 2.899 × 1021 Hg atoms
    3. 4.826 × 1022 atoms of chlorine

12. Write a balanced chemical equation for each reaction and then determine which reactant is in excess.

    1. 2.46 g barium(s) plus 3.89 g bromine(l) in water to give barium bromide
    2. 1.44 g bromine(l) plus 2.42 g potassium iodide(s) in water to give potassium bromide and iodine
    3. 1.852 g of Zn metal plus 3.62 g of sulfuric acid in water to give zinc sulfate and hydrogen gas
    4. 0.147 g of iron metal reacts with 0.924 g of silver acetate in water to give iron(II) acetate and silver metal
    5. 3.142 g of ammonium phosphate reacts with 1.648 g of barium hydroxide in water to give ammonium hydroxide and barium phosphate

13. Under the proper conditions, ammonia and oxygen will react to form dinitrogen monoxide (nitrous oxide, also called laughing gas) and water. Write a balanced chemical equation for this reaction. Determine which reactant is in excess for each combination of reactants.

    1. 24.6 g of ammonia and 21.4 g of oxygen
    2. 3.8 mol of ammonia and 84.2 g of oxygen
    3. 3.6 × 1024 molecules of ammonia and 318 g of oxygen
    4. 2.1 mol of ammonia and 36.4 g of oxygen

14. When a piece of zinc metal is placed in aqueous hydrochloric acid, zinc chloride is produced, and hydrogen gas is evolved. Write a balanced chemical equation for this reaction. Determine which reactant is in excess for each combination of reactants.

    1. 12.5 g of HCl and 7.3 g of Zn
    2. 6.2 mol of HCl and 100 g of Zn
    3. 2.1 × 1023 molecules of Zn and 26.0 g of HCl
    4. 3.1 mol of Zn and 97.4 g of HCl

15. Determine the mass of each reactant needed to give the indicated amount of product. Be sure that the chemical equations are balanced.

    1. NaI(aq) + Cl2(g) → NaCl(aq) + I2(s); 1.0 mol of NaCl
    2. NaCl(aq) + H2SO4(aq) → HCl(g) + Na2SO4(aq); 0.50 mol of HCl
    3. NO2(g) + H2O(l) → HNO2(aq) + HNO3(aq); 1.5 mol of HNO3

16. Determine the mass of each reactant needed to give the indicated amount of product. Be sure that the chemical equations are balanced.

    1. AgNO3(aq) + CaCl2(s) → AgCl(s) + Ca(NO3)2(aq); 1.25 mol of AgCl
    2. Pb(s) + PbO2(s) + H2SO4(aq) → PbSO4(s) + H2O(l); 3.8 g of PbSO4
    3. H3PO4(aq) + MgCO3(s) → Mg3(PO4)2(s) + CO2(g) + H2O(l); 6.41 g of Mg3(PO4)2

17. Determine the percent yield of each reaction. Be sure that the chemical equations are balanced. Assume that any reactants for which amounts are not given are in excess. (The symbol Δ indicates that the reactants are heated.)

    1. KClO3(s)→ΔKCl(s)+O2(g);2.14 g of KClO3 produces 0.67 g of O2
    2. Cu(s) + H2SO4(aq) → CuSO4(aq) + SO2(g) + H2O(l); 4.00 g of copper gives 1.2 g of sulfur dioxide
    3. AgC2H3O2(aq) + Na3PO4(aq) → Ag3PO4(s) + NaC2H3O2(aq); 5.298 g of silver acetate produces 1.583 g of silver phosphate

18. Each step of a four-step reaction has a yield of 95%. What is the percent yield for the overall reaction?

19. A three-step reaction yields of 87% for the first step, 94% for the second, and 55% for the third. What is the percent yield of the overall reaction?

20. Give a general expression relating the theoretical yield (in grams) of product that can be obtained from x grams of B, assuming neither A nor B is limiting.

    A + 3B → 2C

21. Under certain conditions, the reaction of hydrogen with carbon monoxide can produce methanol.

    1. Write a balanced chemical equation for this reaction.
    2. Calculate the percent yield if exactly 200 g of methanol is produced from exactly 300 g of carbon monoxide.

22. Chlorine dioxide is a bleaching agent used in the paper industry. It can be prepared by the following reaction:

    NaClO2(s) + Cl2(g) → ClO2(aq) + NaCl(aq)
    1. What mass of chlorine is needed for the complete reaction of 30.5 g of NaClO2?
    2. Give a general equation for the conversion of x grams of sodium chlorite to chlorine dioxide.

23. The reaction of propane gas (CH3CH2CH3) with chlorine gas (Cl2) produces two monochloride products: CH3CH2CH2Cl and CH3CHClCH3. The first is obtained in a 43% yield and the second in a 57% yield.

    1. If you use 2.78 g of propane gas, how much chlorine gas would you need for the reaction to go to completion?
    2. How many grams of each product could theoretically be obtained from the reaction starting with 2.78 g of propane?
    3. Use the actual percent yield to calculate how many grams of each product would actually be obtained.

24. Protactinium (Pa), a highly toxic metal, is one of the rarest and most expensive elements. The following reaction is one method for preparing protactinium metal under relatively extreme conditions:

    2 PaI 5 (s) → Δ 2 Pa(s) + 5I 2 (s)
    1. Given 15.8 mg of reactant, how many milligrams of protactinium could be synthesized?
    2. If 3.4 mg of Pa was obtained, what was the percent yield of this reaction?
    3. If you obtained 3.4 mg of Pa and the percent yield was 78.6%, how many grams of PaI5 were used in the preparation?

25. Aniline (C6H5NH2) can be produced from chlorobenzene (C6H5Cl) via the following reaction:

    C6H5Cl(l) + 2NH3(g) → C6H5NH2(l) + NH4Cl(s)

    Assume that 20.0 g of chlorobenzene at 92% purity is mixed with 8.30 g of ammonia.

    1. Which is the limiting reactant?
    2. Which reactant is present in excess?
    3. What is the theoretical yield of ammonium chloride in grams?
    4. If 4.78 g of NH4Cl was recovered, what was the percent yield?
    5. Derive a general expression for the theoretical yield of ammonium chloride in terms of grams of chlorobenzene reactant, if ammonia is present in excess.

26. A stoichiometric quantity of chlorine gas is added to an aqueous solution of NaBr to produce an aqueous solution of sodium chloride and liquid bromine. Write the chemical equation for this reaction. Then assume an 89% yield and calculate the mass of chlorine given the following:

    1. 9.36 × 1024 formula units of NaCl
    2. 8.5 × 104 mol of Br2
    3. 3.7 × 108 g of NaCl

1. 1. 53.941 amu
   2. 49.0072 amu
   3. 58.3197 amu
   4. 310.177 amu
   5. 73.891 amu
   6. 81.071 amu

3. 1. 0.3831 mol Si
   2. 4.2 × 10−2 mol Pb
   3. 0.102 mol Mg
   4. 6.8 × 10−3 mol La
   5. 3.78 × 10−2 mol Cl2
   6. 1.2 × 10−3 mol As

5. 1. 9.80 × 10−3 mol or 9.80 mmole Ba(OH)2
   2. 8.08 × 10−3 mol or 8.08 mmole H3PO4
   3. 2.91 × 10−2 mol or 29.1 mmole K2S
   4. 4.634 × 10−3 mol or 4.634 mmole Cu(NO3)2
   5. 1.769 × 10−2 mol 17.69 mmole Ba3(PO4)2
   6. 4.38 × 10−2 mol or 43.8 mmole (NH4)2SO4
   7. 4.06 × 10−3 mol or 4.06 mmole Pb(C2H3O2)2
   8. 1.96 × 10−2 mol or 19.6 mmole CaCl2· 6H2O

7. 1. 613 g or 6.13 × 105 mg Ag
   2. 296 g or 2.96 × 105 mg Sn
   3. 16.6 g or 1.66 × 104 mg Os
   4. 48.9 g or 4.89 × 104 mg Si
   5. 0.764 g or 764 mg H2
   6. 92.05 g or 9.205 × 104 mg Zr

9. 1. 1.81 kg P4O10
   2. 0.387 kg Ba(OH)2
   3. 0.923 kg K3PO4
   4. 0.458 kg Ni(ClO3)2
   5. 0.118 kg (NH4)NO3
   6. 0.109 kg Co(NO3)3

11. 1. 1.91 × 104 mg Pt
    2. 965.6 mg Hg
    3. 2841 mg Cl

13. The balanced chemical equation for this reaction is

    2NH3 + 2O2 → N2O + 3H2O

15. 1. 150 g NaI and 35 g Cl2
    2. 29 g NaCl and 25 g H2SO4
    3. 140 g NO2 and 27 g H2O

23. 1. 2.24 g Cl2
    2. 4.95 g
    3. 2.13 g CH3CH2CH2Cl plus 2.82 g CH3CHClCH3

25. 1. chlorobenzene
    2. ammonia
    3. 8.74 g ammonium chloride.
    4. 55%
    5. Theoretical yield (NH4Cl) = mass of chlorobenzene (g)  ×  0 .92  ×  53 .49 g/mol 112 .55 g/mol