Stoichiometry Chapter 11 & Chapter 13.3

Objectives Determine mole ratios from coefficients in a reaction Predict quantities of reactants and products in chemical reactions Relate moles and gas volumes by stoichiometry (Ch. 13.3) Determine limiting reactants in a chemical reaction Determine excess reactant and how much INXS Determine theoretical, actual and percentage yields

Stoichiometry Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction It is based on the chemical equation (Ch. 9) and the relationship between mass and moles (Ch. 10).

Stoichiometry We can interpret a chemical equation in terms of number of molecules (or ions or formula units) or in terms of number of moles of molecules etc. depending on our needs. Remember that because moles can be converted to mass, we can also produce a mass interpretation of a chemical equation.

Stoichiometry Consider the reaction 2Na + Cl2  2NaCl Molecular interpretation: 2 atom Na + 1 molecule Cl22 Formula Units of NaCl Mole interpretation: 2 mole Na + 1 mole Cl2  2 moles of NaCl Mass Interpretation: 2 mol x 23.0g/mol + 1mol x 71.0 g/mol 2molx58.5 g/mol 46.0 g Na + 71.0 g Cl2  117.0 g 117.0 g = 117.0 g

Examples 2 KBr(s) + Cl2(g) 2KCl(s) + Br2(l) Show conservation of mass by determining the masses of all the reactants and products in the following reactions. 2 KBr(s) + Cl2(g) 2KCl(s) + Br2(l) 238.0 g 71.0g 149.2 g 159.8 g 309 g = 309 g Fe2O3 (s)+ 3CO(g)  2Fe(s) + 3CO2(g) 159.6 g 84.0g 111.6 g 132.0g 243.6 g = 243.6 g

Mole Ratios A mole ratio is ratio between the number of moles of any two substances in a balanced chemical equation. In the reaction 2 Al(s) + 3Br2(l)  2AlBr3(s) What mole ratios can we write? 2 mol Al and 2 mol Al and 3 mol Br2 3 mol Br2 2 mol AlBr3 2 mol AlBr3 3mol Br2 and 2 mol AlBr3 and 2 mol AlBr3 2 mol Al 2 mol Al 3 mol Br2

Mole Ratios These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical Example: How many moles of chlorine are needed to react with 5 moles of sodium (without any sodium left over)? 2Na + Cl2  2NaCl 5 moles Na 1 mol Cl2 2 mol Na = 2.5 moles Cl2

Mole to Mole Conversions Example: If 0.575 mole of CO2 is produced by the combustion of propane, C3H8, how many moles of oxygen are consumed? The balanced equation is: ? 0.575 C3H8 + 5 O2  3 CO2 + 4 H2O ANALYSIS: Relating two compounds usually requires a mole-to-mole ratio SOLUTION:

Mole-Mole Conversion Practice Write the balanced reaction for hydrogen gas reacting with oxygen gas to make water. 2 H2 + O2  2 H2O To make 4.5 moles of water, how many mol of each reactant are needed? If we had 3.2 moles of oxygen, how many mol of hydrogen gas would we need? If we had 52.2 moles of hydrogen, how mol of oxygen gas are needed? If we had made 26.4 mol of water, how many mol of oxygen were used?

Other Conversions Get into groups of not more than 5. Write a procedure or procedures for mass/mole and mass/mass conversions. You can make one procedure for the different types or you can make write two procedures. Follow your procedure EXACTLY as written to demonstrate how your group’s procedure(s) will solve the problems on the next slide.

Mass-Mole Conversion Problems 1. How many moles of hydrogen gas can you make if you react 2.50 g of Na with water? 2. How many grams of aluminum oxide would you need to produce 7.12x103 moles of aluminum metal? 3. If you want to make 62.8 g of Mg3(PO4)2, how many grams each of MgCl2 and Na3PO4 would you need? 4. If we combust 95.6 g of propane (C3H8): A) How many grams of oxygen gas do we need? B) How many grams of CO2 are produced?

Mass-Mass Conversion Problems 1. If we start with 95.6 g of propane (C3H8): A) How many grams of oxygen gas do we need? B) How many grams of CO2 are produced? C3H8 + 5O2  3CO2 + 4H2O 2. If you want to make 62.8 g of Mg3(PO4)2, how many grams each of MgCl2 and Na3PO4 would you need?

Mole-Mass Conversions Most of the time in chemistry, the amounts are given as mass, in grams, instead of moles. We still go through moles and use the mole ratio, but now we also use molar mass to get to grams Example: How many grams of chlorine gas are required to react completely with 5.00 moles of sodium to produce sodium chloride? 2 Na + Cl2  2 NaCl Start with mole then go to mass 5.00 moles Na 1 mol Cl2 70.90g Cl2 2 mol Na 1 mol Cl2 = 177g Cl2

Mole-Mass Conversion Example Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum. 2 Al + 3I2  2AlI3 2 Al + 3I2  2AlI3

Mole – Mass Conversion Practice Amounts of Substance in a Chemical Reaction Find the mass of water produced when 3.68 mol of NH3 is consumed according to the equation below. Assume there is enough O2 present. 3.68 mol ? 4NH3 + 5O2  4 NO + 6H2O

Mass-Mole Conversions Start with mass and convert to moles of product or another reactant Use molar mass and the mole ratio to get to moles of the compound of interest Calculate the number of moles of ethane (C2H6) needed to produce 10.0 g of water 2 C2H6 + 7 O2  4 CO2 + 6 H20 10.0 g H2O 1 mol H2O 2 mol C2H6 18.0 g H2O 6 mol H20 = 0.185 mol C2H6

Mass-Mole Conversion Practice Calculate how many moles of oxygen gas are required to react with aluminum to make 10.0 g of aluminum oxide. 4 Al(s) + 3O2(g) 2 Al2O3 (s) 4 Al + 3O2  2 Al2O3 After this slide give out Chem I Stoichiometry WS 1 - Mole Ratios and Mole to Mole and Mole to Mass Conversions Option: do back of WS 1 as CW/Do Now.

Mass-Mass Conversions Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!) Now we can go from grams to moles, mole ratio, and back to grams of compound we are interested in

Road Map for Mass-Mass conversions

Mass-Mass Conversions Any reaction is going to react in the mole ratio given in the equation. 2Al(s) + 3Br2  2AlBr3(s) So no matter how many moles you have in the reaction, it will react in the 3 mol Br2/2 mol Al ratio. So you have to convert the grams to moles first Example: If you start with 54.0 g of Al, how many g of Br2 do you need? Answer is 479.4g.

Mass-Mass Conversions Example: If you start with 54.0 g of Al, how many g of Br2 do you need? 2Al(s) + 3Br2  2AlBr3(s) 54.0g Al x 1 mol Al x 3 mol Br2 x 159.8 g Br2= 479.4g 27.0 g Al 2 mol Al mol Br2

Mass-Mass Conversions What happens if we start off with 25.0 g of Al? How much Br2 would you need then? You still do the same thing: 25.0g Al x 1 mol Al x 3 mol Br2 x 159.8 g Br2 27 g Al 2 mol Al mol Br2 = 221.9 g Br2 Mol Al = 25.0/27.0 = 0.926 mol Al Mol Br2 = 221.9/159.8 = 1.39 mol Br2

Mole ratios So we have 1.39 mol Br2 and 0.926 mol Al What is the mole ratio of Br2/Al? 1.39 mol Br2/0.926 mol Al 1.5 mol Br2/mol Al Remember that 3 mol Br2/2 mol Al =

Road Map for Mass-Mass conversions

The Steps in a Stoichiometric Mass to Mass Calculation Mass of substance A Use molar mass of A Moles of substance A Use coefficients of A & B in balanced eqn Moles of substance B Use molar mass of B Mass of substance B

Ch. 13.3 – Gas Stoichiometry Remember Avogadro’s Law stated that the volume of a gas at constant pressure and temperature is directly proportional to the number of moles of gas. Volume α n; hold P & T constant V = k x n This means that in a balanced reaction, the coefficients also can represent the ratio of volumes of gas, since equal volumes of gases at the same T & P have the same number of particles.

Gas Stoichiometry C3H8 + 5O2  3CO2 + 4H2O From our reaction from before: C3H8 + 5O2  3CO2 + 4H2O Also means the 1 L of C3H8 reacts with 5 L of O2 to make 3 L of CO2 and 4 L of H2O. Example: What volume of oxygen is needed for the complete combustion of 4.00 L of propane (C3H8)? Assume same T & P. 4.00 L C3H8 x 5 L O2 1 L C3H8 = 20. L O2(g)

Gas Stoichiometry Practice 1 A. Determine the volume of hydrogen gas needed to react completely with 5.00 L of oxygen gas to form water. (Assume both gases at same T & P.) Report answer to 1 decimal place in Quizdom. 2H2(g) + O2(g)  2H2O(l)

Gas Stoichiometry Practice 1 What volume of oxygen gas is needed to completely combust 2.36 L of methane gas (CH4)? Hint: write a balanced reaction first. Input final answer to 3 SF’s in Quizdom.

Gas Stoichiometry Can involve volumes, masses or moles! mass to moles by molar mass (usual) Volume to moles of gas using the ideal gas law (PV =nRT) or 22.4 L/mol Relate mass of solids (or liquids) to volume of gas Remember to convert temperature to Kelvin if given as Celsius.

Gas Stoichiometry Example 1 Ammonia is synthesized by the reaction below. If 5.0 L of nitrogen reacts completely with hydrogen at a pressure of 3.00 atm and 298 K, how many grams of ammonia are formed? N2(g) + 3H2(g)  2NH3(g) Know: V(N2) = 5.0 L; P = 3.00 atm; T = 298 K Ask: What volume of NH3 made under those conditions? 5.0 L N2 x 2 L NH3 1 L N2 = 10.0 L NH3

Gas Stoichiometry Example 1 Now that you know volume, can calculate moles. n= PV RT = 3.00 atm (10.0 L) 0.08206 L∗ atm mol∗k (298K) = 1.23 mol NH3 1.23 mol NH3 x 17 g/mol = 20.8 g NH3

Gas Stoichiometry Example 2 Ammonium nitrate is a common ingredient in chemical fertilizers. Use the reaction below to calculate the mass of solid ammonium nitrate that must be used to obtain 0.100 L of dinitrogen monoxide gas at STP (273 K and 1.0 atm). NH4NO3(s)  N2O(g) + 2H2O(g) Helpful info: Molar mass of NH4NO3 = 80 g/mol “ N2O = 44 g/mol Procedure: Vol gas mol gas  mol NH4NO3  g NH4NO3

Gas Stoichiometry Example 2 NH4NO3(s)  N2O(g) + 2H2O(g) Helpful info: Molar mass of NH4NO3 = 80 g/mol “ N2O = 44 g/mol Procedure: Vol gas mol gas  mol NH4NO3  g NH4NO3 n = PV/RT = 1.0 atm∗0.100L 0.08208 L−atm mol−K ∗(273 K) n = 0.00446 mol N2O* 1 mol NH4NO3 1 mol N2O ∗80 g mol = 0.357 g NH4NO3

Gas Stoichiometry Practice When baking soda (sodium bicarbonate) decomposes, it gives off CO2(g). If we decompose 1.50 g of baking soda, what volume of CO2 gas (in mL) could we collect at 1.00 atm and 298 K? (The molar mass of sodium bicarbonate is 84.0 g/mol). 2NaHCO3(s)  Na2CO3(s) + CO2(g) + H2O(g) 1.50 g * 1 mol/84.0 g * 1 mol CO2/2 mol NaHCO3 = = 0.00893 mol CO2 V = nRT/P = 0.00893 mol*(0.08206 L-atm/mol-K)*298 K/1.00 atm = 0.218 L or 218 mL. Determine answer and enter in Quizdom with 3 SF’s.

Ch. 11.3 - Limiting Reactants Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant. That reactant is said to be in excess (there is too much). The other reactant limits how much product we get. Once it runs out, the reaction s. This is called the limiting reactant.

To Find Limiting Reactant Try all of the reactants. Calculate how much of a product we can get from each of the reactants The reactant that makes the least amount of product is the limiting reactant. Be sure to pick a product! You can’t compare to see which is greater and which is lower unless you compare the results for the same product!

Limiting Reactants The limiting reactant (or limiting reagent) is the reactant that is entirely consumed when a reaction goes to completion. The moles of product are always determined by the starting moles of the limiting reactant.

Which is the limiting reactant? Strategy: Use the relationships from the balanced chemical equation You take each reactant in turn and ask how much product would be obtained, if each were totally consumed. The reactant that gives the smaller amount of product is the limiting reactant.

Limiting Reactant Example 1 If 20 mol hydrogen is reacted with 20 mol of oxygen to form water, which is the limiting reagent? 2H2 + O2 2H2O Only 20 mol of H2O can be formed before H2 runs out!!

Limiting Reactant Example 2 When 100.0 g mercury is reacted with 100.0 g bromine to form mercury (II) bromide, which is the limiting reagent? Hg + Br2 HgBr2

Limiting Reactant Example 2 Moles of HgBr2 produced: From Hg: From Br2: Thus the limiting reagent is How much HgBr2 is made?

Limiting Reactant Practice 1 Iron(III) nitrate reacts with sodium hydroxide to make iron(III) hydroxide. If 0.750 moles of Fe(NO3)3 are added to 2.00 moles of NaOH, how many moles of Fe(OH)3 are made? Fe(NO3)3(aq) + 3NaOH  Fe(OH)3(s) + 3NaNO3(aq) Determine your result and enter into Quizdom with 3 SF’s.

Limiting Reactant Practice 2 15.0 g of potassium reacts with 45.0 g of iodine. Determine which reactant is the limiting reactant and calculate how many grams of product are made. 2K(s) + I2(s) 2KI(s) Determine your result and enter into Quizdom with 3 SF’s.

Limiting Reactant Practice 3 5.00 g of Cu metal react with a solution containing 20.0 g of AgNO3 to produce Ag(s). Determine which reactant is the limiting reactant and calculate how much silver metal is made in grams. Cu(s) + 2AgNO3(aq) 2Ag(s) + Cu(NO3)2(aq) Determine your result and enter into Quizdom with 3 SF’s. Least amount of product made Therefore, AgNO3 is the limiting reactant.

Finding the Amount of Excess By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess. (Note: amount can mean either mole or mass.) Can we find the amount of excess potassium from Practice Problem 1?

Finding Amount of Excess Example - moles We know from Practice Problem 1 that NaOH is LR, so Fe(NO3)3 is in excess. How much is left over in moles? Fe(NO3)3(aq) + 3NaOH  Fe(OH)3(s) + 3NaNO3(aq) 2.00 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 x 1 𝑚𝑜𝑙 𝐹𝑒 𝑁𝑂3 3 3 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 = 0.667 mol Fe(NO3)3 USED! 0.750 mol Fe(NO3)3 give – 0.667 mol used = 0.083 mol Fe(NO3)3 leftover

Finding Amount of Excess Example - mass 15.0 g of potassium reacts with 45.0 g of iodine. 2 K + I2  2 KI We found that Iodine is the limiting reactant, and 58.9 g of potassium iodide are produced. 45.0 g I2 1 mol I2 2 mol K 39.1 g K 253.8 g I2 1 mol I2 1 mol K = 13.9 g K USED! 15.0 g K – 13.9 g K = 1.1 g K EXCESS Given amount of excess reactant Amount of excess reactant actually used Note that we started with the limiting reactant! Once you determine the LR, you should always start with it!

Limiting and Excess Reactants Recap You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT of reactants. Convert ALL of the reactants to the SAME product (pick any product you choose.) The reactant that gave you the lowest amount of product (in either grams or moles) is the LIMITING REACTANT.

Limiting and Excess Reactants Recap The other reactant(s) are in EXCESS. To find the amount of excess, subtract the amount used from the given amount. If you have to find more than one product, be sure to start with the limiting reactant. You don’t have to determine which is the LR over and over again!

Limiting Reactant/Excess Reactant Practice 5.00 g of Cu metal react with a solution containing 20.0 g of AgNO3 to produce Ag(s). We know AgNO3 is the limiting reactant from Practice Problem3. Determine how much copper is leftover in grams. Cu(s) + 2AgNO3(aq) 2Ag(s) + Cu(NO3)2(aq) Enter answer in Quizdom to 3 SF’s.

Theoretical and Actual Yields – Ch. 11.4 Theoretical yield is the amount of product predicted by the stoichiometry. Actual yield is the amount actually obtained in an experiment. Percent yield is the actual yield divided by the theoretical yield. It is a measure of the efficiency of your reaction. (How much bang for the buck!) Percent Yield = Actual Yield x 100 Theoretical Yield Start of 11.4

Theoretical and Actual Yields Example: 0.500g of Mg was burned in excess oxygen and 0.752g of MgO was obtained. Reaction: 2Mg(s) + O2(g)  2MgO(s) What is the theoretical yield? 0.500 g Mg x 1 mol Mg 24. 3 g Mg x 2 mol MgO 2 mol Mg x 40.3 g MgO mol MgO =0.829 g MgO What is the actual yield? What was the percentage yield?

Theoretical and Actual Yield Practice 1 If we burn 8.00 g of CH4, what is the theoretical yield of water? If we only produce 17.0 grams of water, what is the percent yield? (Enter in Quizdom to 1 decimal place.) CH4 + 2O2  CO2 + 2H2O

Theoretical and Actual Yield Practice 2 Ethanol (C2H5OH) is produced from the fermentation of sugar (C12H22O11) in the presence of enzymes. Determine the theoretical and percent yields of ethanol if 684 g of sucrose ferments to form 302 g of ethanol. C12H22O11(aq) + H2O(l)  4C2H5OH(l) + 4CO2(g)