Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Chemistry Study Notes: Chemical Reactions, Stoichiometry, and Redox Reactions, Lecture notes of Chemistry

Unit 4: Chemical Reactions. Significant Numbers Review. Always assume that significant numbers are applied to all problems even if it isn't stated.

Typology: Lecture notes

2021/2022

Uploaded on 09/12/2022

claire67
claire67 🇬🇧

4.6

(5)

265 documents

1 / 5

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Unit 4: Chemical Reactions
Significant Numbers Review
Always assume that significant numbers are applied to all problems even if it isn’t stated.
1) Non-zero digits are always significant numbers
- Ex. 1288 is 4 significant numbers
Zero Significant rules
2) If zero is sandwiched between two non-zero digits, then it is always significant
- Ex. 10002 has 5 significant numbers
3) Leading zeroes (from the left) are never significant
- Ex. 0.004 has only 1 significant number
4) Trailing zeroes (from the right) after a nonzero digit is significant only if there is a
decimal place.
- Ex. 60000. is 5 significant numbers
Stoichiometry
Mass relationships between substances in a chemical relationship
Based on the mole ratio
Indicated by coefficients in a balanced equation
2Mg + 1O2 2MgO
Calculation steps
Write a balanced equation
Identify the known and unknown
What you already know / given information?
What you want to figure out / what’s not given?
Line up Conversion Factors
Mole Ratio - moles to moles
Molar Mass - moles to grams
Molarity - moles to liters solution
Molar Volume - moles to liters gas
*Molar volume at STP: 1 mol of a gas is 22.4 L
pf3
pf4
pf5

Partial preview of the text

Download Chemistry Study Notes: Chemical Reactions, Stoichiometry, and Redox Reactions and more Lecture notes Chemistry in PDF only on Docsity!

Unit 4: Chemical Reactions

Significant Numbers Review

Always assume that significant numbers are applied to all problems even if it isn’t stated.

  1. Non-zero digits are always significant numbers
  • Ex. 1288 is 4 significant numbers Zero Significant rules
  1. If zero is sandwiched between two non-zero digits, then it is always significant
  • Ex. 1 000 2 has 5 significant numbers
  1. Leading zeroes (from the left) are never significant
  • Ex. 0.00 4 has only 1 significant number
  1. Trailing zeroes (from the right) after a nonzero digit is significant only if there is a decimal place.
  • Ex. 60000. is 5 significant numbers

Stoichiometry

● Mass relationships between substances in a chemical relationship ● Based on the mole ratio ○ Indicated by coefficients in a balanced equation ■ 2 Mg + 1 O 2 → 2 MgO ● Calculation steps ○ Write a balanced equation ○ Identify the known and unknown ■ What you already know / given information? ■ What you want to figure out / what’s not given? ○ Line up Conversion Factors ■ Mole Ratio - moles to moles ■ Molar Mass - moles to grams ■ Molarity - moles to liters solution ■ Molar Volume - moles to liters gas ● *Molar volume at STP: 1 mol of a gas is 22.4 L

■ *this is a core step in all stoichiometry problems ○ Example: How many moles of KClO 3 must decompose in order to produce 9 moles of oxygen gas? ■ 2KClO 3 → 2 KCL + 3O 2 ■ 9 moles of O 2 * 2 moles of KClO 3 / 3 moles of O 2 =? ■ Answer = 6 moles of KClO 3 ● Percent Yield ○ Chemical reactions are never perfect ○ You won’t get the amount of product you calculate ○ Actual Yield : Amount of product obtained experimentally ○ Theoretical Yield : Expected amount of product (using stoichiometry) ○ Formula: Actual yield/ Theoretical Yield * 100 ○ Example: What’s the percent yield if 13.1 grams of CaO is actually produced with 24.8 grams of CaCO 3 ■ 24.8 g CaCO 3 * 1 mol CaCO 3 /100.1 g CaCO 3 * 1 mol CaO / 1 mol CaCO 3 * 56.08g CaO/ 1 mol CaO = 13.9 g CaO ■ 13.1/ 13.9 *100 = 94.2% ● Limiting Reagent ○ The reactant that runs out first ○ We care about it bc it determines how much product will be made ● Excess Reactant ○ Added to ensure that the other reactant is completely used up ○ Cheaper and easier to recycle ● How to find the Limiting Reagent? ○ Assume all of the first reactant gets used up ○ Determine how much of one of the products will be made ○ Do the same for the second reactant ○ The reactant that makes less product is the limiting reagent ○ Example: If you have 18 mol SO 2 and 7.0 mol O 2 , what is the limiting reagent? ■ 2SO 2 + O 2 → 2SO 3 ■ 18 mol O 2 *2 mol SO 3 / 2 mol SO 2 = 18 mol SO 3

Oxidation : Loss of electrons but increase in oxidation numbers ● Reduction : Gain of electrons but decrease in oxidation numbers ○ Assigning Oxidation Numbers ■ Elements: ox. number = zero ■ Monatomic Ions: ox. number = charge of ion ■ F in a compound is always - 1 ■ O in a compound is usually - 2 (exception: O in peroxide) ■ H is +1 in covalent bonds ■ Determine other elements from there ● Compounds add up to zero ● Polyatomic Ions add up to their charge ● Redox reactions ○ Identify them when any element’s ox. number changes ○ The element that’s oxidized is the reducing agent and provides the electrons ○ The element reduced is the oxidizing agent ● Balancing Redox Equations ○ “ Half Reaction method,” use when equation is too complicated to balance conventionally ○ If reaction occurs in acid: ■ Write separate half-reactions for oxidation and reduction ■ Balance the elements (except H and O) in each half ■ Balance O using H2O ■ Balance H using H+ ions (available from the acid solution) ○ Balance electrons (e-) in each half (e- are products in oxidation and reactants in reduction) ○ Use multiples of half-reactions (if necessary) to get the numbers of electrons equal ○ Add half-reactions together ○ Add spectator ions back in ● If reaction occurs in base: ○ Write half-reactions and balance H and O as if its an acid

○ To both sides of the balanced equation, add OH- (equal to the number of H+ ions) ○ Form water on the sides that have H+ and OH-. Reduce H2O from both sides if possible ○ Balance the rest as in acid solution