IB DP Chemistry: SL复习笔记1.2.1 Reacting Masses

• The number of moles of a substance can be found by using the following equation:

• It is important to be clear about the type of particle you are referring to when dealing with moles
  • Eg. 1 mole of CaF2 contains one mole of CaF2 formula units, but one mole of Ca2+ and two moles of F- ions

Reacting masses

• The masses of reactants are useful to determine how much of the reactants exactly react with each other to prevent waste
• To calculate the reacting masses, the chemical equation is required
• This equation shows the ratio of moles of all the reactants and products, also called the stoichiometry, of the reaction
• To find the mass of products formed in a reaction the following pieces of information are needed:
  • The mass of the reactants
  • The molar mass of the reactants
  • The balanced equation

Answer:

Step 1: The symbol equation is:

2Mg(s)    +     O₂(g)      →      2MgO(s)

Step 2: The relative atomic masses are:

magnesium : 24.31         oxygen : 16.00

Step 3: Calculate the moles of magnesium used in reaction

Step 4: Find the ratio of magnesium to magnesium oxide using the balanced chemical equation

Therefore, 0.25 mol of MgO is formed

Step 5: Find the mass of magnesium oxide

mass = mol x M

mass = 0.25 mol x 40.31 g mol-1

mass = 10.08 g

Therefore, mass of magnesium oxide produced is 10 g (2 sig figs)

Excess & limiting reactants

• Sometimes, there is an excess of one or more of the reactants (excess reactant)
• The reactant which is not in excess is called the limiting reactant
• To determine which reactant is limiting:
  • The number of moles of the reactants should be calculated
  • The ratio of the reactants shown in the equation should be taken into account eg:

C  + 2H₂      →     CH4

What is limiting when 10 mol of carbon are reacted with 3 mol of hydrogen?

• Hydrogen is the limiting reactant and since the ratio of C : H₂ is 1:2 only 1.5 mol of C will react with 3 mol of H2

Answer:

Step 1: Calculate the moles of each reactant

Step 2: Write the balanced equation and determine the coefficients

2Na + S → Na₂S

Step 3: Divide the moles by the coefficient and determine the limiting reagent

• • divide 0.40 moles of Na by 2, giving 0.20- lowest
  • divide 0.25 moles of S by 1, giving 0.25

Therefore, sodium is limiting and sulfur is in excess