1. Assign oxidation numbers to the "*" marked atoms in the following ions and molecules.
_-1_ Cl in	NaCl*	_-2_ O in	H2O*	_+4__ S in	S*O2
_+3_ N in	N*O21-	_0_ H in	H2*	_+2__ C in	C*O
_+4_ C in	C*O32-	_ +5_ N in	N*O31-	_+4__ C in	C*O2

3.	How many moles of H2 are needed to hydrogenate 10 moles of the unsaturated fatty acid C14H26CO2H and form the saturated fatty acid C14H30CO2H ? Hint balance the equation first. Show work

1. First balance the equation. The coefficients are 1, 2, 1 _1_ C14H25CO2H + __2_ H2 -----> __1_ C14H29CO2H 2. Look at the coefficients to get the numbers for the mole ratios.     3. The mole ratio of unsaturated fatty acid to hydrogen is 1: 2,   1 mole C14H25CO2H / 2 moles H2.     4. This means 1 mole of the unsaturated fatty acid is needed for each 2 moles of hydrogen.     5. When the equation is multiplied by a number like "10" all of the reactants and products must be multiplied by the same number to keep the proportions.   If the reaction uses 10 moles of the fatty acid in reactants then 20 moles hydrogen are needed. __10_ C14H25CO2H + __20_ H2 -----> __10__ C14H29CO2H return to top of the page

4.	What is the number of moles of product formed, when 825 grams of S8(s) reacts with unlimited grams of Cl2(g)?

S8(l) + 4 Cl2(g) -----> 4 S2 Cl2 (g)

This solution is in extreme detail. You would not normally write out everything I have here. I am trying to "overdo" the explanation.

1. Check to be sure the equation is balanced. It is.

2. Identify the mole amounts for the balanced equation.

ideal moles 1 mole S8(l) reacts with 4 moles Cl2(g) -----> 4 moles S2 Cl2 (g) S8(l) + 4 Cl2(g) -----> 4 S2 Cl2 (g)

3. Identify the mole amounts given in the problem. figure molar mass for     1 mole S8(l) = 256 g S8(l)   calculate moles of S8(l) given in the 825 g ; 825 g S8(l)/256 g = 3.22 moles S8(l) This will control the amount of product because the other reactant is available in unlimited amount.

given moles 3.22 moles S8(l) ; unlimited Cl2(g) ---> Unknown moles S2 Cl2 (g) ideal moles from balanced equation 1 mole S8(l) ; 4 moles g Cl2(g) -----> 4 moles S2 Cl2 (g) S8(l) + 4 Cl2(g) -----> 4 S2 Cl2 (g)

4. Determine the multiple for the moles given. The sulfur is the only substance of concern. The chlorine is unlimited and the amount of product is determined by the sulfur.

multiple needed 3.22 times given moles 3.22 moles S8(l) ; and unlimited Cl2(g) ---> Unknown moles S2 Cl2 (g) ideal moles 1 mole S8(l) ; 4 moles g Cl2(g) -----> 4 moles S2 Cl2 (g) S8(l) + 4 Cl2(g) -----> 4 S2 Cl2 (g)

5. The multiple is 3.22. Use this to multiply all reactant and product mole amounts.

actual moles made ---------------------------------------------------------- 4 x 3.22 mols S2 Cl2 (g) actual moles used 3.22 mole S8(l); ....... and........... 4 x 3.22 mols g Cl2(g) given moles 3.22 mole S8(l) ; .....and unlimited Cl2(g) --> Unknown mols S2 Cl2 (g) ideal moles 1 mole S8(l); 4 mols g Cl2(g) -----> 4 mols S2 Cl2 (g) S8(l) + 4 Cl2(g) -----> 4 S2 Cl2 (g)

6. The answers of mols used(needed) and mols made are

actual moles made ---------------------------------------------------------- 12.8 mols S2 Cl2 (g) actual moles used 3.22 mole S8(l) and 12.8 mols g Cl2(g) S8(l) + 4 Cl2(g) -----> 4 S2 Cl2 (g)

5.	Explain why reactions do not typically give 100% yields. What is the percent yield for the production of diborane, B2H6? The predicted yield is 126 g and the actual yield was 81 g? The equation for the reaction is written below.

2 NaBH4(s) + I2(s) ----> B2H6(g) + 2 NaI(s) + H2(g)

A written reaction often has competing reactions. These competing reactions use reactants to make other unwanted products. Also, the reaction may not go to completion. This means the mixture only reaches an equilibrium where reactants remain balanced with products.

Percent yield = 100 [ actual yield/ theoretical yield] here the actual yield is 81 g; the theoretical yield is 126 g; substituting into the definition for percent yield gives

Percent yield = 100 [ 81 g/ 126 g] = 64 % with two significant figures

6.	Phosphoric acid is made in industry by reacting sulfuric acid with mineral phosphates. The reaction between the mineral apatite, Ca5(PO4)3F, and sulfuric acid is written below.

5 H2SO4(aq) + Ca5(PO4)3F + 10 H2O ----> HF + 3 H3 PO4 + CaSO4• 2 H2O

What is the actual yield for the reaction if the theoretical yield is 109 pounds of phosphoric acid, H3 PO4, and the percent yield normally is 83 %?

Percent yield = 100 [ actual yield/ theoretical yield]

83 % = 100 [ actual yield/ 109 pounds] cross multiply by the 109 pounds

[109 pounds] 83 % = 100 [ actual yield/ 109 pounds][109 pounds] result of cross multiplication by the 109 pounds

[109 pounds] 83 %/ 100 = 100/100 [ actual yield] divide by 100

[109 pounds] 0.83 = [ actual yield]

[ actual yield] = [109 pounds] 0.83 = 90. pounds answer has only two significant figures

7.	What is the limiting reagent if 100. moles of hydrogen sulfide, H2S, is mixed with 240. moles of sodium hydroxide, NaOH, according to the following reaction? Show work .

H2S+ 2 NaOH ----> Na2S + 2 H2O

Check to be sure equation is balanced.

Check for ideal mole ratio. The balanced equation gives a mole ratio of 2 mols NaOH : 1 mole H2S

The mole ratio is supposed to be 2 mols NaOH : 1 mole H2S Actual mol ratio is 240 mols NaOH : 100. mole H2S   This is 2.4 mols NaOH to 1 mole H2S and more than the "ideal" ratio.   This means there is more NaOH than needed. The NaOH is in excess. The opposite substance H2S is "limiting" Only 200 mols NaOH needed.   mols NaOH = 100 mols H2S ( 2 mols NaOH /1 mole H2S ) = 200 mols NaOH   if all NaOH reacts then the moles of all reactants and products would be 120 mol H2S 240 mol NaOH 120 mole Na2S 240 mole H2O ----->There are actually only 100 moles of H2S . This is the LIMITING reagent The amounts that can react are limited to the following 100 mol H2S 200 mol NaOH 100 mol Na2S 200 mol H2O This is true because the ratios of moles are 1,2,1,2 as shown here. 1 mol H2S 2 mole NaOH 1 mole Na2S 2 mole H2O H2S+ 2 NaOH ----> Na2S + 2 H2O return to top of the page

8.	What is the limiting reagent if 100. grams of carbon disulfide, CS2, is mixed with 120. grams of sodium hydroxide, NaOH, according to the following reaction? Show work .

3 CS2 + 6 NaOH ----> 2 Na2CS3 + Na2CO3 + 3 H2O

1. Because the question uses mass, the molar masses must be calculated. If the problem only mentions moles then molar masses are not needed. Calculate molar masses for reactants to the nearest gram. The given masses are only good to the nearest gram. there is no need for more significant figures.   1 mol NaOH = 40 g NaOH 1 mol CS2 = 76 g CS2   2. Calculate the numbers of moles of each reactant and compare their ratio to the 'ideal" mol ratio. moles NaOH = (120 g NaOH )( 1 mol NaOH/ 40 g NaOH ) = 3 .0 moles NaOH (2 sf) moles CS2 = (120 g CS2 )/ (76 g CS2 ) = 1.58 moles CS2 3. The "ideal" ratio is 3 moles CS2 /6 moles NaOH or 1 moles CS2 /2 moles NaOH = 0.5   4. The actual ratio is 1.58 moles CS2 / 3 .0 moles NaOH (2 sf) = 0.53 ------>This is more than the ideal There is an excess of carbon disulfide. ------>The sodium hydroxide is the limiting reactant.

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