Chapter 4 Flashcards
In the submicroscopic scale, what the stoichiometric coefficients in the equation refer to ?
atoms, molecules, or formula units
In the macroscopic scale, what the stoichiometric coefficients in the equation refer to ?
the number of moles of each reactant and product.
In stoichiometry, the coefficient in a balanced chemical equation thus allow us to relate __________________________ in a chemical reaction to the ________________________ in that chemical reaction
the amount (moles) of one substance involved
amount (moles) of another substance involved
How do we do stoichiometry calculations in a mole to mole problem?
example: Al + HCl → AlCl3 + H2
How many moles of HCl are needed to react with 0.87 moles of Al?
Step 1: Be sure that the equation is balanced. If not balance it.
not balanced: Al + HCl → AlCl3 + H2
Balanced: 2Al + 6HCl → 2AlCl3 + 3H2
Step 2: find the mole ratio.
2Al:6HCL=2:6
2Al:2AlCl2=2:2
2Al:3H2=2:3
Step 3: Find the moles of the given
0.87 moles of Al will react with HCl
Step 4: Calculate the moles using the ratios
0.87 mol Al X 3 mol HCl/ 1 mol Al
= 2.6 mol HCl
How do we do stoichiometry calculations in a mass to mass problem?
example: Al2O3 → Al + O2
How many grams of Al can be created decomposing 9.8g of Al2O3?
Step 1: Be sure that the equation is balanced. If not balance it.
not balanced: Al2O3 → Al + O2
Balanced: 2Al2O3 → 4Al + 3O2
Step 2: find the mole ratio.
2Al2O3 : 4Al= 2:4
2Al2O3 : 3O2= 2:3
Step 3: Calculate the moles of the given (mol/g)
9.8g Al2O3 x (1mol Al2O3/102g Al2O3) = 0.096 mol Al2O3
Step 4: Calculate the moles using the ratios
0.096 mol Al2O3 x (4 mol Al/2 mol Al2O3) = 0.19 mol Al
Step 5: Calculate the mass using the new moles
0.19 mol Al x (27g Al/1 mol Al) = 5.1g Al
How do we do stoichiometry calculations in a mass-volume problem?
K + H2O→ KOH + H2
How many liters of H2 are created from the reaction of 20.0g K?
knowing that H2 is 22.4L
Step 1: Be sure that the equation is balanced. If not balance it.
Not balanced: K + H2O→ KOH + H2
Balanced: 2K + 2H2O→ 2KOH + H2
Step 2: find the mole ratio.
2K : 2H2O = 2:2
2K : 2KOH = 2:2
2K : 1H2 = 2:1
Step 2: Find the Mass of the Given
20.0g K are used in the reaction
Step 3: Calculate the moles of the given (mol/g)
20.0g K x (1 mol K / 39g K) = 0.513 mol K
Step 4: Calculate the moles using the ratios
0.51 mol K x (1mol H2 /2mol K) = 0.266mol H2
Step 5: Calculate the volume using the new moles
0.266 mol H2 x (22.4L H2 /1mol H2) = 5.75L H2
How do we do stoichiometry calculations in a volume-volume problem?
S2 + O2 → SO2
How many liters of SO2 will be produced from 26.9L O2?
knowing that 1 mol of O2= 22.4L
Step 1: Be sure that the equation is balanced. If not balance it.
Not balanced: S2 + O2 → SO2
Balanced: S2 + 2O2 → 2SO2
Step 2: find the mole ratio.
2O2 : 1S2 = 2:1
2O2 : 2SO2 = 1:1
Step 3: Find the volume of the given
26.9L O2
Step 3: Calculate the moles of the given
26.9L O2 x (1 mol O2 / 22.4L) = 1.20 mol O2
Step 4: Calculate the moles using the ratios
1.20 mol O2 x (1mol SO2 /1mol O2) = 1.20 mol SO2
Step 5: Calculate the volume using the new moles
1.20 mol O2 x (1mol SO2 /1mol O2) x (22.4L /1mol) = 26.9L SO2
What is a limiting reactant?
its amount determines, or limits, the amount of product formed
What are the steps to find the limiting reactant?
example: 2 HCl (aq) + Ca(OH)2 (s) →CaCl2 (aq) + 2 H2O (l)
How many grams of water are made when 0.656 g each of hydrochloric acid (HCl) and calcium hydroxide are mixed?
in the answer:
- Steps 1 and 2 are to find limiting reagent
- Steps 3 to 5 are to find the answer to the previous question
Step 1: Convert the mass of each reactant into moles of product
- 656g HCl X (1 mol HCl / 36.46g of HCL) X ( 2 mol H2O / 2 mol HCl) = 0.0180mol H2O
- 656g Ca(OH)2 X (1 mol Ca(OH)2 / 74.10 g Ca(OH)2) X (2 mol H2O / 1 mol Ca(OH)2) = 0.0177mol H2O
Step 2: Find which reactant produced the less product
- 0180mol H2O are produced with HCl
- 0177mol H2O are produced with Ca(OH)2
So Ca(OH)2 is the limiting reagent HCl is in excess
Step 3: Calculate the mass of product produced
0.0177 mol H2O X ( 18.02 g H2O / 1 mol H2O) = 0.319 g H2O
Step 4: Calculate the moles of excess reactant used and moles that remain.
0.656g HCl X (1 mol HCl / 36.46 g)= 0.0180 mol HCl at start
- 0177 mol H2O X (2 mol HCl / 2 mol H2O)= 0.0177 mol HCl used
- 0180 mol HCl at start – 0.0177 mol HCl used = 0.0003 mol HCl remain
Step 5: Calculate the mass of excess reactant that remain.
0.0003 mol HCl X (36.46 g HCl / 1 mol)= 0.01 g HCl
What is the formula to find the percent yield?
% yield = (actual/theoretical) x 100
How do we determine the empirical formula of a compound by combustion?
The empirical formula of a compound can be determined if the percent composition of the compound is known.
One chemical methods that works well for compounds that burn in oxygen is analysis by combustion.
Each element in the compound combines with oxygen to produce the appropriate oxide
How do we measure concentrations of compounds in solution ?
by molarity:
molarity =(amount of solute x (mol))/(volume of solution (L))
Molarity refers to the amount of _________ per ____
solute per liter
How do we calculate the pH?
PH= -log(H+)
PH= -log(molarity of the acid)
A pH of 7 is ____________
neutral