Solution Stoichiometry

I.  SOLUTIONS

                A. Solutions are homogeneous mixtures composed of a solute and a solvent

                                a. Solute: The chemical present in lesser amount

                                                1. Whatever is being dissolved

                                b. Solvent: The chemical present in greater amount

                                                2. Whatever does the dissolving

                B. Chemicals dissolved in water are aqueous (aq)

II. MOLARITY

                A. Concentration can be expressed in many different ways

                                a. Ex. g/L, mL/L, % by volume, % by mass, mol/L

                                b. The most common is mol/L (M), which is also called molarity

                                                1. mol/L = M

                                                2. [HCl] = find the concentration of HCl

EX.  What is the concentration of 0.118 mol of water in 2.50 L?

0.11 x (1 mol/ 2.50 L) = 0.047M

EX. A solution of Pb(NO₃)₂ has a concentration of 0.330M. How many moles of solute are contained in 100 mL?

(0.330 mol/L) x (0.100L/ 1) = 0.0330 mol

EX. Determine the concentration of 12.5g of PbCl₂ in 30mL of water

12.5g x (1 mol/ 278.2g) x(1/ 0.030L) = 1.5M

EX. What volume of 12.7M  HNO₃ would be needed to produce  25.0g of HNO₃?

(1L/ 12.7 mol) x (25g/ 1) x (1 mol/ 63 g) = 0.031 L

-Nicole

Limiting Reactants

I. In chemical reactions, usually one chemical gets used up first.

                A. This is known as the limiting reactant

                B. Once the L.R. is used up, the reaction stops

                C. The L.R. determines the quantity of products formed

                D. To find the L.R. assume one reactant is used up. Determine how much of this reactant is used up

Ex.

2.5 mol of CuSO₄ reacts with 2.5 mol of NaNO₃. A) Determine the L.R. B)Calculate how many moles of NaNO₃ are formed.

CuSo₄ + 2NaNO₃ --> Cu(NO₃)₂ + Na₂SO₄

A) 2.5 mol x (2/1*) = 5.0 mol of NaNO₃

*What you need over what you have

Since you do not have 5.0 mol of NaNO₃ and you only have 2.5 mol of it, the L.R. is NaNo₃

B) 2.5 mol x (1/2) = 1.25 mol of NaNO₃

Ex.

Determine the L.R. when 1.4 mol of hydrogen reacts with 0.09 mol of oxygen.

2H₂ + O₂ --> 2H₂O

1.4 mol of hydrogen x (1/2) = 0.70 mol of O₂

Therefore the L.R. is H₂

Ex.

What is the L.R. when 125g of P₄ reacts with 320g of Cl₂ to form phosphorus trichloride?

P₄ + 6Cl₂ --> 4PCl₃

125g x (1 mol/ 124g*) x (6/1) x (71.0g/ 1 mol**) = 429g of Cl₂

* Molar mass of P

**Molar mass of Cl

Therefore Cl₂ is the L.R.

Ex. In the formation of water 50g of oxygen react with 60g of hydrogen. Determine the L.R. and theoretical yield.

2H₂ + O₂ --> 2H₂O

50g of O₂ x (1 mol/ 32g) x (2/1) x (2.0g/ 1 mol) = 6.25 of H₂

Therefore the L.R. is O₂

Energy and Percent Yield

I. Enthalpy is the energy stored in chemical bonds

                A. The symbol of enthalpy is H

                                a. units of joules (J)

                B. Change in enthalpy is ΔH

                C. In exothermic reactions enthalpy decreases

                D. In endothermic reactions enthalpy increases

II. Calorimetry

                A. To experimentally determine the heat released we need to know 3 things

                                a. temperature change (ΔT)

                                b.  mass (m)

                                c. specific heat capacity (C)

These are related by the equation:

ΔH = mCΔT

Calculate the heat required to warm a cup of 723 g of water (c=6.98J/g˚C) from 30.0˚C to 50.0˚C
C = 6.98J/g˚C
M = 723g
T = 20.0˚C

▲H = mC▲T
= (723g)(6.98 J/g˚C)(20.0˚C)
= 100930.8
= 1.01×10⁴

Ex. 9500 J of heat are added to a 778 g glass of water initially at 56.0˚C calculate the final temperature of the water (c = 6.21 J/g˚C)
▲H = 9500 J
m = 778 g
C = 6.21 J/g˚C
Ti = 56.0˚C

▲H = mC▲T
= mC (Tf-Ti)

95000J = (778g)(6.21J/g˚C)(Tf-56.0˚C)
95000J = (4831J/˚C)(Tf-56.0˚C)
95000J / 4831J/˚C = (4831J/˚C )(Tf-56.0˚C) / (4831J/˚C )
19.7˚C = Tf-56.0˚C
Tf = 75.7˚C

III. Percent Yield

                A.  The theoretical yield of a reaction is the amount of products that should be formed.

                B. The actual amount depends on the experiment

                c. The percent yield is like a measure of success

                                a. How close is the actual amount to the predicted amount?

Ex. The production of Octane (C₈H₁₈) is given by the equation:
C₈H₁₈ → 8C + 9H₂

If 38.0g of Octane are produced, determine the theoretical (predicted) yield of Carbon.

If 38.0g of Octane are produced, determine the theoretical (predicted) yield of Carbon.

 What is the percent yield of Carbon if 11.0 g is produced?
(11.0 g / 32 g) x 100 = 34%

                

Volume and Heat Conversions

I. Volume @ STP can be found by using the conversion factor 22.4 L/ mol

                A. Heat can be included as a separate term in chemical reactions. (This is called enthalpy)

                                a.  Exothermic reactions are reactions that emit heat

                                b. Endothermic reactions are reactions that absorb heat

                                c. Both can be used in stoichiometry

Ex. When cadmium reacts with hydrochloric acid exactly 3.00L of hydrogen is produced @STP. What mass of cadmium was produced?

Cd + 2HCl --> CdCl₂ + H₂

3.00L x  (1 mol/ 22.4L) x (1 mol/ 1 mol) x (112.4g/ 1 mol*) = 15.1 g

* The 112.4g/ 1 mol is the molar mass of cadmium

Ex. How many litres of chlorine are necessary for the combustion reaction of 362g of selenium @ STP?

Se + Cl₂ -->  SeCl₂

362g x (1 mol/ 79.0g*) x (1 mol/ 1 mol) x (22.4L/ 1 mol) = 103 L

*The 1 mol/ 79.0g is the molar mass of selenium

Ex. Find the amount of heat released when 2.0 mol of O₂ is consumed according to the reaction

Cl₂ + 3O₂ --> 2ClO₃ + 65.9KJ

2.0 mol x (1 mol/ 3 mol) x (65.9KJ/ 1 mol) = 44KJ

-Nicole