Mole-Mole
5C + 2SO2 --> CS2 + 4CO
How many moles of SO2 are required to produce 118.0 mol CO?
5C + 2SO2 --> CS2 + 4CO
118molCO/1 * 2molSO2/4molCO
118molCO/1 * 1molSO2/2molCO = 118molSO2/2 = 59molSO2
Source: Homework #1
Mg + 2HCl --> MgCl2 + H2
How many grams of MgCl2 are produced by 2.55 mol Mg?
Mg + 2HCl --> MgCl2 + H2
2.55molMg/1 * 1molMgCl2/1molMg = 2.55molMgCl2/1 * 95.3g/1mol = 243.0gMgCl2(R)
Source: Homework #1
CaC2 + 2H2O --> C2H2 + Ca(OH)2
If 32.0gCaC2 are used in this reaction, how many moles of calcium hydroxide are produced?
CaC2 + 2H2O --> C2H2 + Ca(OH)2
32gCaC2/1 * 1mol/64.1(R)g = .5(R)molCaC2/1 * 1molCa(OH)2/1molCaC2 = .5molCa(OH)2
Source: Homework #1
_Al(OH)3 + _HCl --> _AlCl3 + _H2O (unbalanced)
If 39.5 g AlCl3 is produced, how many grams of HCl was used in this reaction?
Source: Homework #2
1Al(OH)3 + 3HCl --> 1AlCl3 + 3H2O (balanced)
39.5gAlCl3/1 * 1mol/133.5g(R) = .3molAlCl3/1 * 3molHCl/1molAlCl3 = .9molHCl/1 * 36.5g(R)/1mol = 32.85gHCl
Balance the Equation:
_Zn + _HCl --> _ZnCl2 +_H2
Zn + 2HCl --> ZnCl2 + H2
5C + 2SO2 --> CS2 + 4CO
How many moles of CS2 forms when 2.7 mol C reacts?
5C + 2SO2 --> CS2 + 4CO
2.7molC/1 * 1molCS2/5molC = 2.7molCS2/5 = .54molCS2
Source: Homework #1
Mg + 2HCl --> MgCl2 + H2
If 2.26 moles of HCl are reacted, how many grams of magnesium were used in the reaction?
Mg + 2HCl --> MgCl2 + H2
2.26molHCl/1 * 1molMg/2molHCl = 1.13molMg/1 * 24.3g/1mol = 27.459gMg
Source: Homework #1
CaC2 + 2H2O --> C2H2 + Ca(OH)2
How many moles of water are needed to produce 56.8 g C2H2?
CaC2 + 2H2O --> C2H2 + Ca(OH)2
56.8gC2H2/1 * 1mol/26g(R) = 2.2molC2H2/1 * 2molH2O/1molC2H2 = 4.4molH2O
Source: Homework #1
_Al(OH)3 + _HCl --> _AlCl3 + _H2O (unbalanced)
If this reaction starts with 50.3 g Al(OH)3, how many grams of AlCl3 would you expect to produce?
Source: Homework #2
1Al(OH)3 + 3HCl --> 1AlCl3 + 3H2O (balanced)
50.3gAl(OH)3/1 * 1mol/78g(R) = .64molAl(OH)3/1 * 1molAlCl3/1molAl(OH)3 = .64molAlCl3/1 * 133.5g/1mol = 85.44gAlCl3
_C6H6 + _O2 --> _H2O + _CO2
Balance the Equation.
2C6H6 + 15O2 --> 6H2O + 12CO2
_C3H8 + _O2 --> _CO2 + _H2O (unbalanced)
How many moles of CO2 are produced from the combustion of 6.40 mol C3H8?
1C3H8 + 5O2 --> 3CO2 + 4H2O (balanced)
6.4molC3H8/1 * 3molCO2/1molC3H8 = 6.4molCO2 * 3 = 19.2molCO2
Source: Homework #1
Mg + 2HCl --> MgCl2 + H2
Magnesium reacts with hydrochloric acid. If 7.80 mol HCl reacts, how many grams of MgCl2 will be produced?
Source: Notes
Mg + 2HCl --> MgCl2 + H2
7.8molHCl/1 * 1molMgCl2/2molHCl = 3.9molMgCl2/1 * 95.3g/1mol = 371.7gMgCl2
_N2 + _H2 --> _NH3 Balance the equation.
How many moles of nitrogen gas are required to produce 25.0 g NH3 @ STP?
1N2 + 3H2 --> 2NH3 (Balanced)
25gNH3/1 * 1mol//17g(R) = 1.5molNH3/1 * 1molN2/2molNH3 = .75molN2 @ STP
Source: Homework #2
_NaCl + _H2SO4 --> _HCl + _Na2SO4 (Unbalanced)
What is the mass, in grams, of sodium chloride that reacts with 275.0 g of sulfuric acid?
Source: Stoichiometry Quiz
2NaCl + 1H2SO4 --> 2HCl + 1Na2SO4 (Balanced)
275gH2SO4/1 * 1mol/98g(R) = 2.8molH2SO4/1 * 2molNaCl/1molH2SO4 = 5.6molNaCl/1 * 58.5/1mol = 327.6gNaCl
4NH3 +5O2 --> 6H2O + 4NO
How many liters of NO, at STP are produced from 825.7 g NH3?
Source: Notes
4NH3 +5O2 --> 6H2O + 4NO
825.7gNH3/1 * 1mol/17.03g(R) = 48.5molNH3(R)/1 * 4molNO/4molNH3 = 48.5molNO/1 * 22.4L/1mol =
1086.4 L NO @ STP
_N2 + _H2 --> _NH3 Balance the equation.
How many moles of NH3 form from a reaction of 6.50 mol H2 @ STP?
1N2 + 3H2 --> 2NH3 (Balanced)
6.5molH2/1 * 2molNH3/3molH2 = 13molNH3/2 =
4.3(R) mol NH3 @ STP
Source: Homework #2
_NaCl + _H2SO4 --> _HCl + _Na2SO4 (Unbalanced)
If 12.3 mol HCl are produced in this reaction, how many grams of sodium sulfate are produced?
2NaCl + 1H2SO4 --> 2HCl + 1Na2SO4 (Balanced)
12.3molHCl/1 *1molNa2SO4/2molHCL = 6.15molNa2SO4/1 * 142g/1mol = 873.3gNa2SO4
Source: Stoichiometry Quiz
Double Jeopardy!!! (This will be hard to get)
C6H10O5+H2O-->C6H12O6
If you have 25.3g of starch and 22.0g of dihydrogen monoxide and you end up with 18.044g of glucose, what's the percent yield and how much excess will be left of the excess reactant?
The answer is: 64% and 19.19g of H2O
Ca(OH)2 + Na2SO4 --> CaSO4 + 2NaOH
How many grams of Calcium Sulfate should be produced from 215.0 grams of calcium hydroxide?
Source Stoichiometry Quiz
Ca(OH)2 + Na2SO4 --> CaSO4 + 2NaOH
215gCa(OH)2/1 * 1mol/74.1g(R) = 2.9molCa(OH)2/1 * 1molCaSO4/1molCa(OH)2 = 2.9molCaSO4/1 * 136.1g/1mol = 394.69gCaSO4
_Fe2O3 + _CO --> _Fe + _CO2
If 4.00 kg of iron (III) oxide are reacted, how many moles of carbon monoxide are needed?
1Fe2O3 + 3CO --> 2Fe + 3CO2
4kgFe2O3/1 * 1000/1 = 4000gFe2O3/1 * 1mol/159.6g(R) = 25.1(R)molFe2O3/1 * 3molCO/1molFe2O3 = 75.3molCO
Source: Homework #2
4NH3 + 5O2 --> 6H2O + 4NO
How many moles of water are produced if the reaction begins with 7.50 moles NH3?
4NH3 + 5O2 --> 6H2O + 4NO
7.5molNH3/1 * 6molH2O/4molNH3 =
7.5molNH3/1 * 3molH2O/2molNH3 = 11.3(R)molH2O
Source: Notes
2Al + 3CuSO4 --> Al2(SO4)3 +3Cu
In a single replacement reaction between Al and CuSO4, 2.15 g Al is reacted in an excess of CuSO4 and a 59.5% yield was obtained. What mass, in grams, of copper was produced? (Hint: Find the actual yield)
Source: Slides Percent Yield
2Al + 3CuSO4 --> Al2(SO4)3 +3Cu
2.15gAl/1 * 1mol/27g(R) = .08molAl/1 * 3molCu/2molAl = .12molCu/1 * 63.55g/mol = 7.63gCu %Y = (A/T)*100 --> A = %Y * T/100
.595 * 7.63 = 4.53gCu
What is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data?
Stoichiometry
2Al + 3Cl2 --> 2AlCl3
How many grams of aluminum chloride will be produced if the reaction begins with 135.0 g Al and 75.0 g Cl2
Source: Limiting and Excess Reactant Quiz
135g * 1mol/27g(R) = 5molAl * 3molCl2/2molAl = 7.5mol * 71g(R)/1mol = 532.5gCl2
75g * 1mol/71g(R) = 1.1molCl2 * 2molAl/3molCl2 = .73mol(R)Al * 27g(R)/1mol = 19.71gAl (Use 1.1mol).
1.1molCl2 * 2molAlCl3/3molCl2 = .73mol * 133.5/1 = 97.455gAlCl3
_Al + _ZnCl2 --> _AlCl3 + _Zn Balance the Equation
If 17.8 g Al react, how many moles of each product will be produced?
Source: Homework #2
2Al + 3ZnCl2 --> 2AlCl3 + 3Zn
#1(Zn) 17.8gAl/1 * 1mol/27g(R) = .7molAl(R)/1 * 3molZn/2molAl = 1.05molZn
#2(AlCl3) 17.8gAl/1 * 1mol/27g(R) = .7molAl(R)/1 * 2molAlCl3/2molAl = .7molAlCl3