Home Page Science Engineering Chemical Engineering Balanced Chemical Equation Chapter 4 notes - Flashcards

Chapter 4 notes – Flashcards

Judith Simpson

7 July 2022

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question

what is the equation for stoichiometry?

answer

g --> mol --> mol --> g

question

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?

answer

(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

question

what is a limiting reactant?

answer

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

question

4FeS₂ + 11O₂ --> 2Fe₂O₃ + 8SO₂
(a) how many moles of Fe₂O₃are formed when 2.83mol O₂reacts with excess FeS₂?
(b) how many grams of SO₂are formed when 1.00mol of FeS₂react? (molar mass SO₂ = 64.0g/mol)
(c) how many moles of O₂are required to produce 75.8g of Fe₂O₃? (molar mass Fe₂O₃ = 160.0g/mol)
(d) how many grams of Fe₂O₃are formed when 187g of FeS₂react with excess O₂? (molar mass FeS₂ = 120g/mol)

answer

(a) mol O₂--> mol Fe₂O₃
2.83mol O₂X (2mol Fe₂O₃/11mol O₂) = .515mol

(b) mol FeS₂--> mol SO₂ --> g SO₂
1.00mol FeS₂X (8mol SO₂/4mol FeS₂) X (64.0g SO₂/1mol SO₂) = 128g SO₂

(c) g Fe₂O₃--> mol Fe₂O₃ --> mol O₂
75.8g Fe₂O₃ X (1mol Fe₂O₃/160.0gFe₂O₃) 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

question

the thermite reaction produces iron metal and aluminum oxide from a mixture of powdered aluminum metal and iron (III) oxide.
Fe₂O_3(s) + 2Al_(s) --> 2Fe_(l) + Al₂O_3(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)

answer

(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)
Fe₂O₃is the limiting reactant
(b) g Fe₂O₃ --> mol Fe₂O₃ --> mol Fe --> g Fe
50.0g Fe₂O₃X (1mol Fe₂O₃/159.7g Fe₂O₃) X (2mol Fe/1mol Fe₂O₃) X (55.8g Fe/1mol Fe) = 35.0g Fe

question

the reaction between aluminum metal and chlorine gas produces aluminum chloride as follows:
2Al_(s) + 3Cl_2(g) --> 2AlCl_3(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?

answer

(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

question

what is theoretical yield?

answer

the amount of product that would result if all the limiting reagent reacted

question

what is actual yield?

answer

the amount of product actually obtained from a reaction

question

what is the %yield equation?

answer

%yield = (actual yield/theoretical yield) X 100

question

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)

answer

g FeS₂ --> mol FeS₂ --> mol Fe₂O₃ --> g Fe₂O₃
85.3gFeS₂ 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%

question

what mass of potassium iodide id required to make 500.mL of a 2.80M KI solution?
(KI = 166.0g/mol)

answer

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

question

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 CO₃^2- ions?
(Na₂CO₃ = 106.0g/mol)

answer

M = (mol solute/L solution)
M = [25.3g Na₂CO₃ X (1mol Na₂CO₃/106.0g Na₂CO₃)]/0.250L
M = 0.955mol/L
Na₂CO_3(s) --> 2Na⁺_(aq) + CO₃^2-_(aq)
.955M 2(.955M) .955M
[Na⁺] = 1.91M [CO₃^2-] = .955M

question

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

answer

M_iV_i = M_fV_f
M_i = 4.00M V_i = ?
M_f= 0.200M V_f= 60.0mL

V_i = (M_fV_f/M_i)
V_i = (0.200M X 60mL/ 4.00M)
V_i = 3mL

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

question

solution stoichiometry

answer

g        g
↓         ↑
mol → mol
↑         ↓
L        L

question

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

Zn_(s) + 2HCl_(aq) --> ZnCl_2(aq) + H_2(g)

answer

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

question

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

Na₂CO_3(aq) + 2HNO_3(aq) --> 2NaNO_3(aq) + CO_2(g) +
H₂O_(l)

(Na₂CO₃ = 106.0g/mol)

answer

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

question

what is a titration?

answer

a solution of accuratelty known concentration (standard solution) is added gradually to another solution of unknown concentration to determine its concentration

question

what is the equivalence point?

answer

the point in a titration when stoichiometric amounts of acid and base have been combined
Ex: moles H⁺ = moles OH^-

question

what is an indicator?

answer

a substance that changes color at (or near) the equivalence point, indicating the end of the reaction

question

what is an endpoint?

answer

the point in a titration when the indicator changes color

question

what is a standard solution?

answer

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)

question

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)

answer

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

question

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) --> C₈H₄O₄^2-_(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)

answer

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

question

how many grams of KHP are needed to neutralize 18.64mL of 0.1004M NaOH solution?
HC₈H₄O₄^-_(aq) + OH^-_(aq) --> C₈H₄O₄^2-_(aq) + H₂O_(l)

(KHC₈H₄O₄ = 204.2g/mol)

answer

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

Chapter 4 notes – Flashcards

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