Bell Work: Limiting Reactant Problems You have 8 car bodies and 28 tires. As you build cars, which will you run out of first? Which will limit the number of complete cars produced? Which do you have excess of? How much excess do you have? How many running cars will be produced?

You have 8 car bodies (Cb) and 28 tires (Tr). Write a reaction: 1 Cb + 4 Tr  1 Cr As you build cars, which will you run out of first? Tires 2. Which will limit the number of complete cars produced? tires 3. Which do you have excess of? Car bodies How much excess do you have? 28 𝑇𝑟 1 𝐶𝑏 4 𝑇𝑟 =7 Cb used up. So, 8 -7 = 1 leftover car body. An excess of 1 car body. 5. How many running cars will be produced? 7

Vocabulary Limiting reagent: chemical first used up in a reaction Produces least amount of product Limits amount of product that can be produced The reaction will stop when all of the limiting reactant is consumed Excess reagent: chemical not used up in a reaction The excess reactant remains because there is nothing with which it can react. How much is left over? Take the amount you start with minus how much reacted with the limiting reagent. Percent yield= actual yield theoretical yield x 100.

Limiting Reactant Example 1 To make 12 cookies, you need 2 cups of sugar and 3 cups of flour. If you have 8 cups of sugar and 8 cups of flour, which is the limiting reagent? 2 S + 3 F  12 C 8 𝑐𝑢𝑝𝑠 𝑠𝑢𝑔𝑎𝑟 12 𝑐𝑜𝑜𝑘𝑖𝑒𝑠 2 𝑐𝑢𝑝𝑠 𝑠𝑢𝑔𝑎𝑟 =48 𝑐𝑜𝑜𝑘𝑖𝑒𝑠 8 𝑐𝑢𝑝𝑠 𝑓𝑙𝑜𝑢𝑟 12 𝑐𝑜𝑜𝑘𝑖𝑒𝑠 3 𝑐𝑢𝑝𝑠 𝑓𝑙𝑜𝑢𝑟 =32 𝑐𝑜𝑜𝑘𝑖𝑒𝑠 Since the flour will make fewer cookies, it is the limiting reagent.

2 S + 3 F  12 C 2. Which do you have excess of? Sugar 3. How much excess do you have? 8 𝑐𝑢𝑝𝑠 𝑓𝑙𝑜𝑢𝑟 2 𝑐𝑢𝑝𝑠 𝑠𝑢𝑔𝑎𝑟 3 𝑐𝑢𝑝𝑠 𝑓𝑙𝑜𝑢𝑟 =5.33 cups sugar are needed for the recipe. So there are 8 – 5.33 = 2.67 cups of excess sugar. How many cookies will be produced? 8 𝑐𝑢𝑝𝑠 𝑓𝑙𝑜𝑢𝑟 12 𝑐𝑜𝑜𝑘𝑖𝑒𝑠 3 𝑐𝑢𝑝𝑠 𝑓𝑙𝑜𝑢𝑟 =32 cookies

Limiting Reactant Example 2 2 Al + 3 I2 ------> 2 AlI3 a. Determine the limiting reagent and the theoretical yield of the product if one starts with 1.20 mol Al and 2.40 mol iodine. 1.2 𝑚𝑜𝑙 𝐴𝑙 2 𝑚𝑜𝑙 𝐴𝑙 𝐼 3 2 𝑚𝑜𝑙 𝐴𝑙 = 1.2 mol Al I3 2.4 𝑚𝑜𝑙 𝐼 3 2 𝑚𝑜𝑙 𝐴𝑙 𝐼 3 3 𝑚𝑜𝑙 𝐼 3 = 1.6 mol Al I3 Aluminum is the limiting reagent.

2 Al + 3 I2 ------> 2 AlI3 b. If 0.9 mol were actually produced in the lab, what is the percent yield of the reaction? 0.9 1.2 𝑥100=75% c. Can you ever have a yield greater than 100%? No! Mass would not be conserved.

How many moles of carbon dioxide will be produced? Example 3 2 C5H10 + 15 O2  10 CO2 + 10 H2O If 25 grams of pentane react with 25 grams of oxygen, what is the limiting reactant? How many moles of carbon dioxide will be produced? How much excess reactant will remain after the reaction? If 20.0 g were actually produced, what was the percent yield?

Example 3 2 C5H10 + 15 O2  10 CO2 + 10 H2O If 25.0 grams of pentane react with 25.0 grams of oxygen, what is the limiting reactant? 25 𝑔 C5H10 𝟏 𝒎𝒐𝒍C5H10 𝟕𝟎 𝒈C5H10 x 10 𝑚𝑜𝑙 𝐶𝑂2 𝟐 𝒎𝒐𝒍C5H10 =1.79 𝑚𝑜𝑙 𝐶𝑂2 25 𝑔 O2 𝟏 𝒎𝒐𝒍O2 𝟑𝟐 𝒈O2 x 10 𝑚𝑜𝑙 𝐶𝑂2 𝟏𝟓 𝒎𝒐𝒍O2 =0.521 𝑚𝑜𝑙 𝐶𝑂2 Oxygen is limiting because it produced less CO2.

2 C5H10 + 15 O2  10 CO2 + 10 H2O How many moles of carbon dioxide will be produced? 0.521 𝑚𝑜𝑙 𝐶𝑂2 How much excess reactant will remain after the reaction? 0.521 𝑚𝑜𝑙 𝐶𝑂2 2 𝑚𝑜𝑙 𝐶5𝐻10 10 𝑚𝑜𝑙 𝐶𝑂2 𝑥 70 𝑔𝐶5𝐻10 1𝑚𝑜𝑙 𝐶5𝐻10 =7.29 𝑔 𝑢𝑠𝑒𝑑 𝑢𝑝 25.0 – 7.29 = 17.71 g pentane left over

Theoretical yield in grams: 0.521 𝑚𝑜𝑙 𝐶𝑂2 44 𝑔 𝐶𝑂2 1 𝑚𝑜𝑙 𝐶𝑂2 =22.9 g Example 3 2 C5H10 + 15 O2  10 CO2 + 10 H2O If 20.0 g of carbon dioxide were actually produced, what was the percent yield? Theoretical yield in grams: 0.521 𝑚𝑜𝑙 𝐶𝑂2 44 𝑔 𝐶𝑂2 1 𝑚𝑜𝑙 𝐶𝑂2 =22.9 g percent yield: 20.0 22.9 𝑥100=87%