Unit 1 Part 2 CALCULATIONS AND CHEMICAL EQUATIONS

I Know the Answer! Tee Hee!

Topics:  Stoichiometry ◦ definition – what it is ◦ types of stoichiometry ◦ mole ratios ◦ calculations using n & m ◦ other calculations using v & VSTP   

Theoretical yield, actual yield and percent yield Limiting and Excess Reagent Core Lab 2 – % Yield in a Chemical Reaction



Stoichiometry comes from the Greek words stoikheion, meaning “element” and metron, meaning “to measure”.

 Stoichiometry

is the determination of quantities needed for, or quantities produced by, chemical reactions.

3 types of stoichiometry 1. Gravimetric stoichiometry  stoichiometry using mass (g) of solids(s). 2. Solution stoichiometry  stoichiometry using concentration & volume of solutions. 3. Gas stoichiometry  stoichiometry using volume (L) of gases (g) at STP.

Q: How do we calculate quantities used or quantities produced in chemical reactions?

A:

Mole Ratios

•The coefficients from a balanced chemical equation provide the mole ratios of reactants and products in a chemical reaction. •Coefficients from a balanced chemical equation, work like a sandwich recipe . . .

Clubhouse sandwich recipe

MOLE RATIOS Fill in the missing quantities: Slices of Toast

Slices of Turkey

Strips of Bacon

27

66 100

# of Sandwiches 12



Mole ratios from balanced chemical equations work in the same way

eg: equation for an air bag exploding sodium azide  sodium + nitrogen gas

2 NaN3(s)  2 Na(s) + 2 units or 2 moles

NOT 2 g

3 N2(g)

2 NaN3(s)



2 Na(s)

+

3 N2(g)

6

10 150 10

MOLE RATIOS 



A mole ratio is a mathematical expression that shows the relative amounts of two species involved in a chemical change. A mole ratio ◦ comes from a balanced chemical equation ◦ show the relative amounts of reactants & products in moles

◦ looks like

coefficien t required coefficien t given 9:10 AM



Mole ratios from balanced chemical equations represent relative number of particles (atoms, molecules, ions, formula units or moles).

C3H8(g) +

5 O2(g)

 3 CO2(g) + 4 H2O(g)

15 mol 18 molecules 35 molecules 7.38 mol

NOTE: Mole ratios DO NOT REPRESENT MASS. P4 + 5 O2 → 2 P2O5 1 g of phosphorus reacts with 5 g of oxygen

STEPS: 1. Write a balanced chemical equation with subscripts (Identify given and required values). 2. Use MOLE RATIOS from balanced chemical equation .

nrequired = ngiven x

coefficien t required coefficien t given

eg.1 How many moles of ammonia gas are produced when 0.60 mol of nitrogen gas reacts with hydrogen gas? N2 + H2 → NH3

(1.2 mol)

eg.2 How many moles of CO2 are produced when 31.5 mol of C3H8 is burned? C 3H 8 +

O2 → CO2 + H2O

(94.5 mol)

eg.3 Calculate the number of moles of magnesium needed to produce 0.586 mol of Mg3N2. (1.76)

eg.4 How many moles of sulfur are produced when 4.25 mol of SO3 decomposes? (0.531) 8 SO3 → 1 S8 + 12 O2 4.25 mol ? nS8 = 4.25 mol SO3 x 1 mol S8 8 mol SO3

p. 114 #’s 2 & 3 p. 115 #’s 4 – 7 p. 117 #’s 8 - 10

Stoichiometry #1



Other types of Stoichiometry: ◦ Mole to Mass (or Mass to Mole) ◦ Mass to Mass ◦ Volume to Volume (for gases at STP) ◦ SolutionStoichiometry (Part 3 of Unit 1...)



Mole to Mass Calculations are the same as Mole to Mole but also require conversion between mass and moles:



Use

n = m M

or

m = nxM

STEPS: 1. Balanced chemical - identify given and required 2. Convert mass to moles 3. Mole Ratio

nrequired = ngiven x

coefficien t required coefficien t given

eg. 1: Calculate the number of moles of oxygen that will react with 6.49 g of aluminum. 4 Al(s) + 3 O2(g) → 2 Al2O3(s) 6.49 g

n=

6.49 26.98

?? mol

= 0.2405 𝑚𝑜𝑙

nO2 = 0.2405 mol x

3 𝑚𝑜𝑙 𝑂2 4 𝑚𝑜𝑙 𝐴𝑙

= 0.180 mol O2

eg. 2: How many moles of water are produced when 20.6 g of CH4 burns? (2.57 mol)

CH4 + 2 O2 → CO2 + 2 H2O 20.6 g ? mol n = 20.6 g = 1.283 mol CH4 16.05 g/mol

n = 1.283 mol CH4 x 2 H2O 1 CH4 = 2.57 mol H2O

Stoichiometry #2

eg. 3: Calculate the mass of aluminum oxide that can be produced from the reaction between 0.1804 mol of oxygen gas and excess aluminum.

Al(s)

+

0.1804 mol

O2(g) →

Al2O3(s) ?? g

eg. 4: What mass of Ca3P2 is produced when 4.38 mol of Ca reacts with phosphorus? (266 g) 6 Ca (s) + P4 (s) → 2 Ca3P2 (s) 4.38 mol ?g n = 4.38 mol Ca x 2 Ca3P2 6 Ca = 1.46 mol Ca3P2

m = (1.46 mol)(182.18 g/mol) = 266 g Ca3P2

p. 149 #’s 6 & 7 Stoichiometry #3

STEPS: 1. Balanced chemical (given!! required!! subscripts!!) 2. Convert mass to moles. 3. Mole Ratio

nrequired = ngiven x 4.

coefficien t required coefficien t given

Convert moles to mass.

NOTE: The coefficients from balanced

equations represent MOLE RATIOS.

MASSES are NOT to be used in the MOLE RATIO step.

eg1: How many grams of NH3 gas are produced when 5.40 g of hydrogen gas reacts with nitrogen gas?

N2(g) + H2(g) → NH3(g)

eg 2: What mass of nitrogen gas is needed to react completely with hydrogen gas and produce 30.6 g of ammonia gas? (1.796 mol – 0.8979 mol - 25.2 g)

eg 3: What mass of CaCl2(aq) is produced when Ca(NO3)2(aq) reacts with 4.39 g of NaCl(s)? Ca(NO3)2(aq) + 2 NaCl(s) → CaCl2(aq) + 2 NaNO3(aq)

4.39 g 0.07512 mol 0.03756 mol 4.17 g

?g

eg 4: Given that 45 g of sulfur (S8) gas reacts with aluminum solid, find the mass of product. 3 S8 (s) + 16 Al(s) → 8 Al2S3(s) 45 g ?g

0.1754 mol 0.4677 mol 70.2 g = 70. g

Page 121

#’s 11 to 13 & #14

Stoichiometry #4

Stoichiometry #5

LAW OF CONSERVATION OF MASS 

Mass is conserved in chemical reaction.

OR total mass of the reactants 

=

total mass of the products.

We can verify this law by using the mole ratios and molar masses of reactants and products from balanced chemical equations.

Show Mass Conserved ______________ QUANTITY

2 Mg(s) + O2(g)

n (mol)

2

1

m (g)

2(24.31)

32.00



2 MgO(s) 2

1.

The decomposition of 500.00 g of Na3N produces 323.20 g of N2. How much Na is produced in this decomposition?

2.

What mass of oxygen is needed to react with 6.49 g of aluminum to produce 12.26 g of Al2O3?

3.

To produce 90.1 g of water, what mass of hydrogen gas is needed to react with 80.0 g of oxygen?

4.

If 3.55 g of Cl2 reacts with exactly 2.29 g of sodium, what mass of NaCl will be produced?

If gas volume at STP is GIVEN or REQUIRED you must use VSTP in your calculations. eg 1: What volume of oxygen at STP is needed to react with 2.34 g of NO2 ? 2 N2O5(s)  4 NO2(g) + O2(g)

eg 2: Calculate the mass of N2H4(g) needed to produce 157 L of nitrogen gas at STP in the reaction below. 2 N2H4(l) + N2O4(l)  3 N2(g) + 4 H2O(l)

p. 123 # 16, 17

Sandwich Time Again!

Here’s what we find in our kitchen: ◦ How many sandwiches can we make??

◦ What limits the # of sandwiches??

Chemical reactions with 2 or more reactants, stop when 1 reactant is completely used up.  Limiting reagent ◦ AKA limiting reactant ◦ The reactant that is completely used in a chemical reaction, thereby limiting the amount of product that can form. 



Excess reagent ◦ AKA excess reactant ◦ the reactant that is left over after the reaction is complete. ◦ there is more than is needed for complete reaction.



We can predict which reactant is the LR by using mole ratios from balanced chemical equations.

How to Solve: 1. Write balanced chemical equation. 2. Calculate n for EACH reactant. 3. Use the mole ratio for each reactant. The LR is the reactant that produces less moles of product 4. Calculate the amount of product using the LR.

eg 1: Determine the Limiting Reagent and Excess Reagent if 14.8 g of C3H8(g) reacts with 2.14 g of O2(g). C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) 14.8 g 2.14 g ▼ ▼ 0.3355 mol 0.06688 mol ▼ ▼ 1.007 mol CO2 OR 0.04013 mol CO2 O2 is the LR → produces less CO2

eg 2: Identify the excess and limiting reactants when 2.00 g of NaI(aq) reacts with 2.00 g of Pb(NO3)2(aq). Pb(NO3)2(aq) + 2 NaI(aq) → PbI2(s) + 2 NaNO3(aq)

p. 131

#’s 23 - 26

When you are given 2 or more reactants in a stoichiometry question, YOU MUST ALWAYS determine the LR and use it to determine the amount of product.

eg 1. If 6.50 mol of CaBr2 solution reacts with 3.75 mol of NaNO3 solution, then how many moles of Ca(NO3)2 will be produced? CaBr2(aq) + NaNO3(aq)  NaBr(aq) + Ca(NO3)2(aq)

eg 2. What mass of NaCl is produced when 6.70 mol of Na reacts with 3.20 mol of Cl2? Na(s) +

Cl2(g) 

NaCl(s)

eg 3. What mass of CaCO3 will be produced when 20.0 g of Ca3(PO4)2 reacts with 15.0 g of Na2CO3 ? Ca3(PO4)2 +

Na2CO3 →

CaCO3 +

Na3PO4

eg 4. What mass of Ba(OH)2will be produced when 10.0 g of Ba(NO3)2reacts with 30.0 g of NaOH? NaOH + Ba(NO3)2 → NaNO3 +

Ba(OH)2

eg 5. What volume of H2 at STP will be produced when 10.0 g of Zn reacts with 20.0 g of HCl(aq)?

Zn(s) + 2 HCl(aq)  H2(g) + ZnCl2(aq)

p. 134 #’s 27.a), b), 28.a), b) & 30.a)



The term YIELD refers to the mass of product formed in a chemical reaction.

eg. Burning 99.5 g of propane in sufficient oxygen produces 298 g of carbon dioxide gas. THEORETICAL YIELD  what you are supposed to get  The amount of product predicted or calculated.



A question that asks you to find the theoretical yield is asking you to calculate the amount expected.

Actual yield  what you actually get in an experiment  The amount actually produced in a chemical reaction in the LAB.

For most reactions, your actual yield is less than your theoretical yield, for many reasons, including: ◦ Reaction is slow and was incomplete. ◦ Impurities are present. ◦ Reaction goes to equilibrium (never really finishes). ◦ Sample is wet when weighed. ◦ Experimental error.

eg 1. Calculate the theoretical yield of copper if 1.87 g of Al reacts with excess aqueous copper (II) sulfate? 2 Al + 3 CuSO4 → Al2(SO4)3 + 3 Cu 1.87 g Al ►0.06931 mol Al ►0.1040 mol Cu ▼ 6.61 g Cu Theoretical Yield

eg 1.(cont’d) If the reaction produces 3.74 g of copper, what is the percent yield of copper? % yield =

3.74 6.61

x 100 = 56.6 %

eg 2. 20.0 g of bromic acid, HBrO3 (aq) , is reacted with excess HBr. HBrO3(aq) + 5 HBr(aq)  3 H2O(l) + 3 Br2(aq) (a) What is the theoretical yield of Br2 for this reaction? (74.4 g) (b) If 47.3 g of Br2 are produced, what is the percentage yield of Br2? (63.6%)

eg 3. Barium sulfate forms a precipitate in this reaction: Ba(NO3)2 (aq) + Na2SO4 (aq)  BaSO4(s) + 2 NaNO3(aq) When 35.0 g of Ba(NO3)2 reacts with excess Na2SO4, 29.8 g of BaSO4 are recovered by the chemist. (a) Calculate the theoretical yield of BaSO4. (31.3 g)

(b) Calculate the percentage yield of BaSO4. (95.2 %)

eg 4. Yeasts can act on a sugar, such as glucose, C6H12O6, to produce ethyl alcohol, C2H5OH, and carbon dioxide. C6H12O6 (s)  2 C2H5OH(l) + 2CO2(g) If 223 g of ethyl alcohol are recovered after 1.63 kg of glucose react, what is the percentage yield of the reaction? (834 g, 26.7 %)

PERCENT

YIELD lab!