Calculating Amounts of Reactants and Products and The Limiting Reactant Concept

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Lecture #9. Calculating Amounts of Reactants and Products and The Limiting Reactant Concept. Chemistry 142 B James B. Callis, Instructor Autumn Quarter, 2004. Chemical Equation Calculations. Mass. Atoms (Molecules). Molecular Weight. Avogadro’s Number. g/mol. 6.02 x 10 23. Molecules.

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Lecture #9

### Calculating Amounts of Reactants and ProductsandThe Limiting ReactantConcept

Chemistry 142 B

James B. Callis, Instructor

Autumn Quarter, 2004

Chemical Equation Calculations

Mass

Atoms (Molecules)

Molecular

Weight

Number

g/mol

6.02 x 1023

Molecules

Reactants

Products

Moles

Problem 9-1: Calculating Reactants and

Products in a Chemical Reaction -I

Problem: Given the following chemical reaction between aluminum

sulfide and water, if we are given 65.80 g of Al2S3: a) How many moles

of water are required for the reaction? b) What mass of H2S & Al(OH)3

would be formed?

Al2S3 (s) + 6 H2O (l) 2 Al(OH)3 (s) + 3 H2S (g)

Plan: Calculate moles of aluminum sulfide using its molar mass, then

from the equation, calculate the moles of water, and then the moles of

hydrogen sulfide, and finally the mass of hydrogen sulfide using

it’s molecular weight.

Solution:

a) molar mass of aluminum sulfide =

moles Al2S3 =

Calculating Reactants and Products in a

Chemical Reaction - II

a) cont.

H2O: 0.4382 moles Al2S3 x

b)H2S: 0.4382 moles Al2S3 x

___ moles H2O

1 mole Al2S3

___ moles H2S

1 mole Al2S3

molar mass of H2S =

mass H2S =

Al(OH)3: 0.4382 moles Al2S3 x

molar mass of Al(OH)3 =

mass Al(OH)3 =

Calculating the Amounts of Reactants and

Products in a Reaction Sequence - I

Problem 9-2: Calcium Phosphate could be prepared in the following

reactionsequence:

4 P4 (s) + 10 KClO3 (s) 4 P4O10 (s) + 10 KCl (s)

P4O10 (s) + 6 H2O (l) 4 H3PO4 (aq)

2 H3PO4 (aq) + 3 Ca(OH)2 (aq) 6 H2O(aq) + Ca3(PO4)2 (s)

Given: 15.5 g P4 and sufficient KClO3 , H2O and Ca(OH)2. What mass

of Calcium Phosphate could be formed?

Plan: (1) Calculate moles of P4.

(2) Use molar ratios to get moles of Ca3(PO4)2.

(3) Convert the moles of product back into mass by using the

molar mass of Calcium Phosphate.

Calculating the Amounts of Reactants and

Products in a Reaction Sequence - II

Solution:

moles of P4 =

For Reaction #1 [ 4 P4 + 10 KClO4 4 P4O10 + 10 KCl ]

For Reaction #2 [ 1 P4O10 + 6 H2O4 H3PO4 ]

For Reaction #3 [ 2 H3PO4 + 3 Ca(OH)2 1 Ca3(PO4)2 + 6 H2O]

moles Ca3(PO4)2 =

Calculating the Amounts of Reactants and

Products in a Reaction Sequence - III

Molar mass of Ca3(PO4)2 = 310.18 g mole

mass of product =

Balanced

reaction!

Defines

stoichiometric

ratios!

Unbalanced (i.e., non-stoichiometric)

mixture!

Limited by syrup!

Limiting Reactant

In a chemical reaction where arbitrary amounts of reactants are mixed and allowed to react, the one that is used up first is the limiting reactant. A portion of the other reactants remains.

There is a systematic procedure for finding the limiting reagent based on the reactant ratio (RR) defined as the ratio of the number of moles of a reactant to its coefficient in a balanced chemical equation. The reagent with the smallest reactant ratio is the limiting reactant.

For a Reaction of the Form:

aA + bB = cC + dD

If compounds A and B are present in the mole amounts called for in the balanced reaction, then the following equation is valid:

If the Reactant Ratios of all reactants

are equal, then there is no limiting

reactant, and all reactants will be

consumed.

If all reactant ratios are not equal, then the reactant with the smallest reactant ratio is the limiting reactant.

From among these, choose (RR)min, the smallest reactant ratio. This identifies the limiting reactant.

Once (RR)min is identified, it can be used to calculate the actual moles of other reactants and products actually consumed, e.g.

Actual moles of B consumed = (RR)minb

Actual moles of C produced = (RR)minc

Problem 9-3: Acid - Metal Limiting Reactant - I

• 2Al(s) + 6HCl(g) 2AlCl3(s) + 3H2(g)

Consider the reaction above. If we react 30.0 g Al and 20.0 g HCl, how many moles of aluminum chloride will be formed?

• 30.0 g Al
• 20.0g HCl
• Limiting reactant = reactant with RRmin =

Problem 9-3: Acid - Metal Limiting Reactant - II

Calculate the yield of AlCl3 from:

Moles AlCl3 = (RR)min x reaction coefficient of AlCl3.

Problem 9-4: Ostwald Process Limiting Reactant

What mass of NO could be formed by the reaction 30.0g of ammonia gas and 40.0g of oxygen gas w/ the rxn below? 4NH3 (g) + 5 O2 (g) 4NO(g) + 6 H2O(g)

30.0g NH3

40.0g O2

RRmin =_______, therefore ______ is the Limiting Reactant!

Moles NO formed =

Mass NO =

Actual Yield (mass or moles)

Theoretical Yield (mass or moles)

% Yield = x 100%

Chemical Reactions in Practice: Theoretical,

Actual, and Percent Yields

Theoretical yield: The amount of product indicated by the

stoichiometrically equivalent molar ratio in the balanced equation.

Side Reactions: These form different products that

take away from the theoretical yield of the main product.

Actual yield: The amount of product that is actually obtained.

Percent yield (%Yield):

or Actual yield = Theoretical yield x (% Yield / 100%)

Problem 9-5: Percent Yield

Actual Yield

Theoretical Yield

Percent Yield = x 100% =

Problem: The chemical reaction between iron and water to form

the iron oxide, Fe3O4 and hydrogen gas is given below. If 4.55 g of iron is

reacted with sufficient water to react all of the iron to form rust, what is

the percent yield if only 6.02 g of the oxide are formed?

Plan: Calculate the theoretical yield and use it to calculate the percent

yield, using the actual yield.

Solution:

3 Fe(s) + 4 H2O(l) Fe3O4 (s) + 4 H2 (g)

Moles Fe =

Theoretical moles Fe3O4 =

Theoretical mass Fe3O4 =

Problem 9-6: Percent Yield / Limiting Reactant - I

Problem: Ammonia is produced by the Haber process using nitrogen

and hydrogen Gas. If 85.90g of nitrogen are reacted with

21.66 g hydrogen and the reaction yielded 98.67 g of

ammonia what was the percent yield of the reaction.

N2 (g) + 3 H2 (g) 2 NH3 (g)

Plan: Since we are given the masses of both reactants, this is a limiting

reactant problem. First determine which is the limiting reactant

then calculate the theoretical yield, and then the percent yield.

Solution:

moles N2 =

moles H2 =

Problem 9-6: Percent Yield / Limiting Reactant - II

N2 (g) + 3 H2 (g) 2 NH3 (g)

Solution Cont.

We have 3.066 moles of Nitrogen, and it is limiting, therefore the

theoretical yield of ammonia is:

mol NH3 =

mass NH3 =

Actual Yield

Theoretical Yield

Percent Yield = x 100%

98.67 g NH3

g NH3

Percent Yield = x 100% =

Answers to Problems in Lecture # 9
• (a) 2.629 moles H2O; (b) 44.81g H2S, 68.36g Al(OH)3
• 77.61 g Ca3(PO4)2
• 0.183 mol of AlCl3
• 30.0 g NO
• 95.6 %
• 94.49 %