The Easiest Way to Understand Chemistry. Chemistry Concepts, Problems and Solutions. Sergey D Skudaev

produces 100 g of CaCO3

      5.3 g of Na2CO3 produces X g of CaCO3

      X= 5.3 * 100 / 106 = 5 g

      Acid Base reactions

      Molarity. Molality. Normality

      1. Let say we have 100 ml of H2SO4 solution and it contains 0.49 g of H2SO4

      What is Molarity?

      What is Molality?

      What is Normality?

      1 Molar solution contains a number of grams equal to molecular mass per one liter of solution.

      MW of H2SO4 = 2 +32 +4* 16 = 98g

      I mole = 98g/L

      We have to find how many grams of H2SO4 given solution are contained in 1 liter.

      100 ml = 0.1 L and contains 0.49 g

      1 L contains X g

      X = 0.49 * 1 / 0.1 = 4.9 g.

      98g/L – 1 mole

      4.9 g/L – X moles

      X = 4.9 * 1 / 98 = 0.05 moles

      The molarity of a 100 ml solution of H2SO4, which contains 0.49 g of H2SO4, equals 0.05 moles.

      What is Molality? Molality is moles of solute / kg of solvent.

      A solute is a substance dissolved in another substance.

      A solvent is a substance in which another substance is dissolved

      What is normality? An equivalent is the molecular mass or mass of acid or base that produce one mole of protons (H+) or one mole of hydroxyl

      (OH-) ions.

      One mole of H2SO4 produces 2 moles of H+ then equivalent to H2SO4 = MW/2 = 49g/L

      49g/L is 1 Normal solution

      4.9g/L solution is – X N

      X= 4.9 * 1/ 49 = 0.10 N.

      The normality of 100 ml solution of H2SO4, which contains 0.49 g of H2SO4, equals 0.10 N.

      2. We have 10 ml of NaOH unknown concentration. The solution was titrated with 0.10 N solution of H2SO4 and 15 ml were required for neutralization. What is the concentration of NaOH?

      Nb * Vb = Na * Va

      where V is volume, N is Normality, b – base and a – acid

      Nb = Na * Va / Vb = 0.10 * 15 / 10 = 0.15 N

      The concentration of NaOH = 0.15 N.

      Weight and Volume problems:

      1. How many liters of Hydrogen are produced from one liter of water?

      2H20 = 2H2 +02

      First, we have to find how many grams of water are spent and how many grams of H2 are produced?

      MW of H20 = 2 +16 = 18

      If we spent 2 molecules of H20 then we spent 18 * 2 =36 g.

      MW H2 = 2 and we got 2 molecules of H2. 2*2 = 4g

      36 g of water produces 4 g of Hydrogen

      1000 g of water produces X g of Hydrogen.

      X = 4 * 100 / 36 = 111.1 g

      I mole of Gas under normal conditions occupies 22.4 liters.

      So we have to know how many moles of H2 are produced.

      1 mole of H2 equals 2 g

      X mole of H2 equals 111.1g

      X = 111.1g / 2g = 50.6 moles of H2

      1 mole of H2 occupies 22.4 liters

      50.6 moles of H2 occupy X liters

      X = 50.6 moles * 22.4L = 1232 L

      Equilibrium. Le Chatelier’s Principle

      Any chemical reaction goes both ways

      According to Le Chatelier, if at equilibrium point we make any change in concentration, pressure or temperature the point of equilibrium will move to counteract the change.

      If the reaction produces heat then heating the system will move the equilibrium to the left and cooling the system will move the equilibrium to the right. If volume of the products is greater than the volume of the reactants then increasing the pressure will move the equilibrium to the left. Decreasing the pressure will move the equilibrium to the right. Increasing concentration of reactants will move the equilibrium to the right, increasing concentration of the products will move the equilibrium to the left.

      Equilibrium constant K c = [C] * [D] / [A] * [B]

      where [] is concentration in moles

      H2 + I2 <=> 2HI

      The concentration of products and reactants is raised to the power of their respective coefficients.

      K c = [HI] ^2 / [H2] [I2]

      The concentration of products and reactants is raised to the power of their respective coefficients: a, b, c, d.

      Let us calculate the equilibrium constant for the following reaction.

      2Al +6HCl + H20= 2AlCl3 +3H2

      The concentration of HCl is 0,5 M, the concentration of AlCl3 is 0.2M and the concentration of H2 is 0.2M at equilibrium point.