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Chapter 4 notes
ch 4 (chem 1)
Undergraduate 1

Additional Chemistry Flashcards




what is the equation for stoichiometry?
g --> mol --> mol --> g

methanol burns in air according to the equation

2CH3OH + 3O2 --> 2CO2 + 4H2O

If 209g of methanol are used up in the combustion, what mass of water is produced?


(CH3OH = 32.0g/mol) (H2O = 18.0g/mol)

g CH3OH --> mol CH3OH --> mol H2O --> g H2O


209g CH3OH X (1mol CH3OH/32.0g CH3OH) X (4 mol H2O/2 mol CH3OH) X (18.0g H2O/1mol H2O) = 235g H2O

what is a limiting reactant?
the reactant used up first in a chemical reaction and therefore limits the amount of product that can be formed

4FeS2 + 11O2 --> 2Fe2O3 + 8SO2

(a) how many moles of Fe2O3 are formed when 2.83mol O2 reacts with excess FeS2?

(b) how many grams of SO2 are formed when 1.00mol of FeS2 react? (molar mass SO2 = 64.0g/mol)

(c) how many moles of O2 are required to produce 75.8g of Fe2O3? (molar mass Fe2O3 = 160.0g/mol)

(d) how many grams of Fe2O3 are formed when 187g of FeS2 react with excess O2? (molar mass FeS2 = 120g/mol)


(a)  mol O2 --> mol Fe2O3

2.83mol O2 X (2mol Fe2O3/11mol O2) = .515mol


(b) mol FeS2 --> mol SO2 --> g SO2

1.00mol FeS2 X (8mol SO2/4mol FeS2) X (64.0g SO2/1mol SO2) = 128g SO2


(c) g Fe2O3 --> mol Fe2O3 --> mol O2

75.8g Fe2O3 X (1mol Fe2O3/160.0g Fe2O3) X (11mol O2/2mol Fe2O3) = 2.61mol O2


(d) g FeS2 --> mol FeS2  --> mol Fe2O3 --> g Fe2O3

187g FeS2 X (1mol FeS2/120g FeS2) X (2mol Fe2O3/4mol FeS2) X (160g Fe2O3/1mol Fe2O3) = 125g Fe2O3


the thermite reaction produces iron metal and aluminum oxide from a mixture of powdered aluminum metal and iron (III) oxide.

Fe2O3(s) + 2Al(s) --> 2Fe(l) + Al2O3(s)

in one process, 50.0g of Al are reacted with 50.0g of Fe2O3

(a) which one is the limiting reactant?

(b) what mass of iron metal can be produced?

(Al = 26.98g/mol) (Fe2O3 = 159.7g/mol) (Al2O3 = 102.0g/mol) (Fe = 55.8g/mol)


(a) mol Al: 50.0g X (1mol Al/26.98g Al) = 1.85mol Al

     mol Fe2O3: 50.0g X (1mol Fe2O3/159.7g Fe2O3) = .313mol Fe2O3

(2mol Al/1mol Fe2O3)

(1.85mol Al/.313mol Fe2O3) = (5.92mol Al/1mol Fe2O3)

Fe2O3 is the limiting reactant

(b) g Fe2O3 --> mol Fe2O3 --> mol Fe --> g Fe

50.0g Fe2O3 X (1mol Fe2O3/159.7g Fe2O3) X (2mol Fe/1mol Fe2O3) X (55.8g Fe/1mol Fe) = 35.0g Fe


the reaction between aluminum metal and chlorine gas produces aluminum chloride as follows:

2Al(s) + 3Cl2(g) --> 2AlCl3(s)

if 2.70g of alumminum is reacted with 4.05g of chlorine:

(a) which is the limiting reactant?

(b) what mass of AlCl3 can be produced?

(c) what mass of the excess reactant remains after the limiting ractant has been consumed?


(a) mol Al: 2.70g Al X (1mol Al/27.0g Al) = .1mol Al

     mol Cl2: 4.05g Cl2 X (1mol Cl2/71.0g Cl2) = .057mol Cl2

Cl2 is the limiting reactant and Al is in excess

(b) g Cl2 --> mol Cl2 --> mol AlCl3 --> g AlCl3

4.05g Cl2 X (1mol Cl2/71.0g Cl2) X (2mol AlCl3/3mol Cl2) X (133.5g AlCl3/1mol AlCl3) = 5.07g AlCl3

(c) g Cl2 --> mol Cl2 --> mol Al --> g Al

4.05g Cl2 X (1mol Cl2/71.0g Cl2) X (2mol Al/3mol Cl2) X (27.0g Al/1mol Al) = 1.03g Al

2.70g Al - 1.03g Al = 1.67g Al remains

what is theoretical yield?
the amount of product that would result if all the limiting reagent reacted
what is actual yield?
the amount of product actually obtained from a reaction
what is the %yield equation?
%yield = (actual yield/theoretical yield) X 100

if 30.7g of Fe2O3 is obtained from the reaction of 85.3g of FeS2 with excess oxygen, what is the percent yield?

4FeS2 + 11O2 --> 2Fe2O3 + 8SO2

(FeS2 = 120g/mol) (Fe2O3 = 160g/mol)


g FeS2 --> mol FeS2 --> mol Fe2O3 --> g Fe2O3

85.3g FeS2 X (1mol FeS2/120g FeS2) X (2mol Fe2O3/4mol FeS2) X (160g Fe2O3/1mol Fe2O3) = 56.9g Fe2O3


%yield = (30.7g/56.9g) X 100 = 54.0%


what mass of potassium iodide id required to make 500.mL of a 2.80M KI solution?

(KI = 166.0g/mol)


M = mol/L     mol = M x L

mol = (2.80mol/L) X 0.5L = 1.4mol KI

1.40mol KI X (166.0g KI/1mol KI) = 232g KI


if 25.3g of sodium carbonate, Na2CO3, is dissolved in enought water to make 250.mL of solution, what is the concentration of Na2CO3? What are the concentrations of the Na+ and CO32- ions?

(Na2CO3 = 106.0g/mol)


M = (mol solute/L solution)

M = [25.3g Na2CO3 X (1mol Na2CO3/106.0g Na2CO3)]/0.250L

M = 0.955mol/L

Na2CO3(s) --> 2Na+(aq) + CO32-(aq)

.955M           2(.955M)    .955M

[Na+] = 1.91M [CO32-] = .955M

how would you prepare 60.0mL of 0.200M HNO3 from a stock solution of 4.00M HNO3?

MiVi = MfVf

Mi = 4.00M                   Vi = ?

Mf = 0.200M                 Vf = 60.0mL


Vi = (MfVf/Mi)

Vi = (0.200M X 60mL/ 4.00M)

Vi = 3mL


3mL of acid + 57mL of water to make 60mL of solution

solution stoichiometry

  g        g

  ↓         ↑

mol → mol

  ↑         ↓

  L        L


what volume of 2.50M HCl, in milliliters, is required to react completely with 11.8g of Zn?


Zn(s) + 2HCl(aq) --> ZnCl2(aq) + H2(g)


g Zn --> mol Zn --> mol HCl --> L HCl

11.8g Zn X (1mol Zn/65.39g Zn) X (2mol HCl/1mol Zn) X (1L HCl/2.5mol HCl) = 0.144L HCl

=144mL HCl


what mass of Na2CO3, in grams, is required for complete reaction with 50.0mL of 0.125M HNO3?


Na2CO3(aq) + 2HNO3(aq) --> 2NaNO3(aq) + CO2(g) +



(Na2CO3 = 106.0g/mol)


L HNO3 --> mol HNO3 --> mol Na2CO3 --> g NaCO3

.05L HNO3 X (0.125mol HNO3/1L HNO3) X (1mol NaNO3/2mol HNO3) X (106g NaNO3/1mol NaNO3) = 0.331g Na2CO3

what is a titration?
a solution of accuratelty known concentration (standard solution) is added gradually to another solution of unknown concentration to determine its concentration
what is the equivalence point?

the point in a titration when stoichiometric amounts of acid and base have been combined

Ex: moles H+ = moles OH-

what is an indicator?
a substance that changes color at (or near) the equivalence point, indicating the end of the reaction
what is an endpoint?
the point in a titration when the indicator changes color
what is a standard solution?

a solution whose concentration is known to a high degree of accuracy. The standard base that is usually used is NaOH and it is standardized against KHP.

(KPH = KHC8H4O4)


what volume of 0.812M HCl, in milliliters, is required to titrate 1.45g of NaOH to the equivalence point?


HCl(aq) + NaOH(aq) --> NaCl(aq) + H2O(l)


g NaOH --> mol NaOH --> mol HCl --> L HCl

1.45g NaOH X (1mol NaOH/40.0g NaOH)  (1mol HCl/1mol NaOH) X (1L HCl/.812mol HCl) = .0446L HCl

= 44.6mL HCl


potassium hydrogen phthalate (KHP), KHC8H4O4, is used to standardize solutions of bases. the acidic anion reacts with strong bases according to the following net ionic equation:

HC8H4O4-(aq) + OH-(aq) --> C8H4O42-(aq) + H2O(l)

if a 0.902g sample of potassium hydrogen phthalate is dissolved in water and titrated to the equivalence point with 26.45mL of NaOH, what is the molar concentration of the NaOH?


(KHC8H4O4 = 204.22g/mol)


g KHP --> mol KHP --> mol NaOH

.902g KHP X (1mol KHP/204.22g KHP) X (1mol NaOH/1mol KHP) = .002442mol NaOH


M = (.004442mol NaOH/.02645L NaOH)

= 0.167M


how many grams of KHP are needed to neutralize 18.64mL of 0.1004M NaOH solution?

HC8H4O4-(aq) + OH-(aq) --> C8H4O42-(aq) + H2O(l)


(KHC8H4O4 = 204.2g/mol)


L NaOH --> mol NaOH --> mol KHP --> g KHP

.01864L NaOH X (.1004mol NaOH/1L NaOH) X (1mol KHP/1mol NaOH) X (204.2g KHP/1mol KHP) = 0.3822g KHP

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