Warmup1. Calculate the molar mass of SbI3

502.46 g/mole2. According to the equation below, how many

moles of aluminum oxide can be made from 6.8 moles aluminum?

Al + O2 Al2O3

6.8 moles Al moles Al2O3 =

moles Al

4 3 2

4

2 3.4 moles Al2O3

a. What is Avogadro's favorite kind of music?b. What line from Shakespeare do high school moles have to memorize?c. What element do moles love to study in chemistry?remolte control

remoletly sophmole

Rock 'N' Mole

"To mole or not to mole, this is the question.“

Molybdenum

A device used by moles to watch Bill Nye the Science Guy and the Discovery Channel.

Obscurely having to do with a mole.

Anyone in the tenth grade who is taking Chemistry already!

Conversion flow chart

Gramsgiven

Molesgiven

Molesunknown

Gramsunknown

Molarmassstep

Molarmassstep

Mole-moleconversion

step

Limiting Reactants Notes

Problem: If you start with X grams of Givenium, how many grams of Unknownium will be made?

X grams of given (1 mole given )( B moles unknown)(D grams of unknown) =

(A grams of given) ( C moles given ) ( 1 mole unknown )

Key:Given = the compound whose amount is given to you in the problemUnknown = compound whose amount you are looking forX = grams that the problem gives youA = molar mass of the compound given in the problem (see periodic table)B and C = numbers from molar ratio, found in the chemical equationD = molar mass of the unknown compound from the problem (see periodic

table)

Ex 1: How many grams of Al2O3 can be formed from 5.81 grams of Al?

Al + O2 Al2O3

1. Write/Balance Equation:

2. Figure out Given: 5.81 grams Al Figure out Unknown: ? grams Al2O3 3. Choose correct pathway:

g Al mol Al mol Al2O3 g Al2O3

4. Numbers!

5.81 g Al ( g Al2O3 ) ( mol Al)

( mol Al2O3 )

( g Al) ( mol Al) ( mol Al2O3 )

1

26.98

2

4 1

101.96

= 11.0 grams Al2O3

4 3 2

Ex 2: Hydrogen gas and chlorine gas react to form hydrochloric acid (HCl). How many grams of the product will form in the reaction if 3.75 grams of chlorine gas is used?

H2 + Cl2 HCl

2. Figure out Given: 3.75 grams Cl2 Figure out Unknown: ? grams HCl 3. Choose correct pathway:

g Cl2 mol Cl2 mol HCl g HCl

4. Numbers! 3.75 g

Cl2( mol Cl2 ) ( mol HCl )

( g Cl2 ) ( mol Cl2 ) ( mol HCl )

1

70.9

2

1 1

36.46

= 3.86 grams HCl

( g HCl)

1. Write/Balance Equation:

11 2

2Sb(s) + 3I2(s) → 2SbI3(sEx 3. What mass of antimony(III) iodide can form from 1.20g Sb ?

1.20 g Sb ( mol Sb ) ( mol SbI3)

( g Sb ) ( mol Sb) ( mol SbI3 )

( g SbI3 ) 502.46

121.76

1

1

2

2

= 4.94 g SbI3

Wait! We know how much Sb we have. But what if there is just a tiny, tiny amount of

iodine available? Hmmmm…Limiting Reactant – reactant that is used up first in a reaction; it limits how much product you can make!!!

2Sb(s) + 3I2(s) → 2SbI3(s)Ex 4. Determine the theoretical yield (mass in grams) of antimony(III) iodide formed when 1.20g Sb and 2.40 g I2 are mixed.

5. Do another calculation with the other amount given 6. Choose the lower amount…the higher amount is not possible. The theoretical yield is calculated based upon the limiting reactant.

2.40 g I2 ( mol I2 ) ( mol SbI3)

( g I2 ) ( mol I2) ( mol SbI3 )

( g SbI3 ) 502.46

1253.8

1 2

3

= 3.17 g SbI3 wait but 1.20 g Sb predicts 4.94 g SbI3

3.17 g SbI3. I2 is the limiting reactant (even though there is a greater amount of it) and is completely used up. Some Sb will be

leftover.

Ex 5. Suppose 36.0 grams of ammonia and 50.0 grams of oxygen react to form nitrogen gas and water. Calculate the mass of nitrogen gas that could be formed.

NH3 + O2 N2 + H2O34 62

36.0 g NH3( mol NH3 ) ( mol N2)

( g NH3) ( mol NH3)

( g N2 )

50.0 g O2 ( mol O2 ) ( mol N2)

( g O2 ) ( mol O2)

( g N2 )

2

4

2

3

( mol N2 )

( mol N2 )

17.04

1

32.00

1

28.02

1

28.02

1

= 29.2 g N2

= 29.6 g N2

Ex 6. From the previous problem, what mass of excess reactant remains?

NH3 + O2 N2 + H2O

29.2 g N2 were produced. Oxygen gas = limiting reactant. Rephrase this question:

how much ammonia is required to produce 29.2 g N2 ?

34 62

29.2 g N2 ( mol N2 ) ( mol NH3)

( g N2 ) ( mol N2)

( g NH3 )

( mol NH3 )

17.04

12

4

28.02

1

= 35.5 g NH3

35.5 g were actually required but 36.0 grams NH3 were used in the reaction. Therefore, the excess NH3

= 0.5 g

percent yield = actual yield x 100

theoretical yield

Ex 7. In the previous problem, we calculated that 29.2 g N2 should be produced. If 26.7 grams N2 are actually produced, calculate the percent yield of the experiment

percent yield = 26.7 g x 100 29.2 g

91.4 %

Conversion flow chart

Gramsgiven

Molesgiven

Molesunknown

Gramsunknown

Molarmassstep

Molarmassstep

Mole-moleconversion

step

#molecules /atoms /ions given

# unknown molecules / atoms/ions

Ex 8: 6.7 x 1017 molecules chlorine gas can form what mass of hydrochloric acid? Assume an excess of hydrogen gas.

H2 + Cl2 HCl

Pathway: molec. Cl2 mol Cl2 mol HCl g HCl

6.7 x 1017 molec Cl2

( mol Cl2 ) ( mol HCl )

( molec. Cl2 ) ( mol Cl2 ) ( mol HCl )

1

6.02x 1023

2

1 1

36.46

= 8.1 x 10-5 grams HCl

( g HCl)

11 2

Ex 9: 98.7 g Cl2 will completely react with how many moles of hydrogen gas?

H2 + Cl2 HCl

Pathway: g Cl2 mol Cl2 mol H2

98.7 g Cl2 ( mol Cl2 ) ( mol H2 )

( g Cl2 ) ( mol Cl2 )

1 1

1

= 1.39 moles H2

11 2

70.9