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

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.

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)

(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) + 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?

(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

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 CO₃^2- 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

Na₂CO_3(s) --> 2Na⁺_(aq) + CO₃^2-_(aq)

.955M           2(.955M)    .955M

[Na⁺] = 1.91M [CO₃^2-] = .955M

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

  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) --> ZnCl_2(aq) + H_2(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?

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

H₂O_(l)

(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

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

Ex: moles H⁺ = moles OH^-

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) --> 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)

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) --> C₈H₄O₄^2-_(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