11 Stoichiometry Calculations Based On Chemical Equations
11 Stoichiometry Calculations Based On Chemical Equations
11 Stoichiometry Calculations Based On Chemical Equations
Stoichiometry
Stoichiometry is the calculation of quantities of substances involved in chemical reactions. Balanced chemical equations, through their coefficients indicate the ratio of moles of reactants and products:
N2
1 mol N2 3 mol H2 1 mol N2 2 mol NH3 3 mol H2 2 mol NH3
3H2
or 3 mol H2 1 mol N2
2NH3
Mole ratios:
or
or
Mole-Mole Calculations
Given the moles of one reactant or product, calculate the moles required or produced in the other reactant(s) or product(s). Steps:
1. Write the balanced chemical equation. 2. Write the moles of the given reactant or product. 3. Multiply by the mole ratio: (moles desired)/(moles starting chemical)
+ 3H2
3 molecules of hydrogen (each containing 2 atoms) to form:
2NH3
2 molecules of ammonia. ( Each molecule contains 1 atom of nitrogen and 3 atoms of hydrogen.)
N2
3H2
2NH3
Calculate the moles of NH3 that can be produced from 10.8 moles of H2. Mole C = mol A x (c/a) Mol NH3 = mol H2 x [(coef NH3)/(coef H2)] mol NH3 = 10.8 mol H2 [( 2 NH3)/(3 H2)] Mol NH3 = 7.20 mol NH3
bB
X (Molar Mass B)
Example Problem
N2 + 3 H2 2NH3 Start with 50 g N2. How many grams of NH3 will be produced? MM N2 = 2 x 14 = 28 g MM NH3 = 14 + 3x1 = 17 g mol N2 = g/MM = 50/28 = 1.78 mol mol NH3 = 1.78 mol N2 x (2 NH3/1 N2)= 3.56 mol g NH3 = mol x MM = 3.56x 17 = 60.53 g
Moles to Grams
In doing a Stoichiometry calculation, you can start and end at any point in the process, depending upon what is given and what is desired: Grams A moles A moles B grams B To Convert from moles A to grams B, follow the sequence: moles A moles B grams B
N2 + 3 H2
2NH3
Calculate the grams of N2 that will react with 12.0 moles of H2. Moles N2 = (12.0 moles H2)x(1 N2/3 H2) = 4.00 moles N2 Molar Mass N2 = 2(14.0) = 28.0 g/mol grams N2 = 4.00 moles x 28.0 g/mol = 112 g N2
Limiting Reactants
In a chemical reaction, there is an exact ratio of reactants, determined by the balanced chemical equation. If there is too much of one of the reactants, it is said to be in excess. The other reactant is said to be the limiting reactant (or reagent).
Example LR problem
N2 + 3H2 2 NH3 If you start with 50.0g of Nitrogen and 45.0g of Hydrogen, Determine which reactant is in Excess and which is Limiting. Moles N2 = 50/28 = 1.78 Moles H2 = 45/2 = 22.5 Divide by coeff. 1.78/1 = 1.78 ; 22.5/3 =7.5 nitrogen is less,: LR, hydrogen is in excess.
Calculate theoretical, actual and percentage yield, given appropriate data. The theoretical yield would be the amount of product you would predict to be produced from a given quantity of reactant(s) from stoichiometric calculations. The actual yield is the amount of product actually obtained by weighing. %yield = (actual/theoretical) x 100%
Example Problem
When 5.00g of N2 are reacted with 7.00g of H2, 5.08g of NH3 are produced. Calculate the theoretical and the percentage yield. The actual yield is 5.08g. The balanced equation is: N2(g) + 3H2(g) 2NH3(g) 1. Determine the Limiting Reagent:
a. Determine the moles of each reactant. Mol = g/M; M(N2) = 2x14 = 28
When 5.00g of N2 are reacted with 7.00g of H2, 5.08g of NH3 are produced. Calculate the theoretical and the percentage yield. N2(g) + 3H2(g) 2NH3(g)
M(H2) = 2x1 = 2 mol N2 = 5.00/28 = 0.179 mol mol H2 = 7.00/2 = 3.5 mol b. Divide each of the mol by its coefficient:
N2: 0.179/1 = 0.179 H2: 3.5/3 = 1.17
c. Choose the smaller of the answers in b. as the Limiting Reagent: N2 : Base further calculations on LR: N2
When 5.00g of N2 are reacted with 7.00g of H2, 5.08g of NH3 are produced. Calculate the theoretical and the percentage yield.
N2(g) + 3H2(g) 2NH3(g) 2. Calculate theoretical yield based on the LR, N2:
a. Mol N2 = 0.179 mol (see previous calc.) b. Mol NH3 = mol N2x (coef. NH3/coef. N2) = 0.179x(2/1) = 0.358 mol c. mass NH3 = molxM M = 14 + 3(1) = 17 mass NH3 = 0.358 x 17 = 6.09 g = Th. Yield
Percentage Yield
Percentage Yield = (Actual/Theoretical)x100 = (5.08/6.09)x100 = 83.4%
The heat energy given up may be treated as a product: In the combustion of methane:
CH4(g) + 2 O2(g) CO2(g) + H2O(g) + heat
Enthalpy
The amount of energy given off in a chemical reaction is equal to the difference in chemical energy between the reactants and products. When pressure is constant, this energy change is called the enthalpy of the reaction, (H. The reaction of hydrogen with oxygen to form water is exothermic:
H2(g) + O2(g) H2(g) + O2(g) H2O(l) + 283 kJ H2O(l) (,!kJ
Endothermic Reactions
The reverse of this reaction corresponds to the electrolysis of water.
H2O(l) + 283 kJ H2(g) + O2(g) H2O(l) H2(g) + O2(g) (, = +283 kJ
Endothermic Reactions have a positive enthalpy change. The reverse reaction has an enthalpy change of the same magnitude but opposite sign.
Quantitative Considerations
H2(g) + O2(g) H2O(l) (,!kJ Calculate the heat given off when 5.0 g of water is produced: The heat is proportional to the number of moles. The above reaction is for 1 mol. 5.0 g/ 18g/mol = 0.277 mol heat given off = (-283 kJ/mol) x 0.277mol= 78.6 kJ
Endothermic reactions can occur because systems also tend to become more random.
This randomness is called entropy. Change in entropy is (S. According to the Second Law of Thermodynamics, the entropy of the universe is always increasing. If in a chemical reaction the entropy of the universe increases, the reaction is spontaneous.