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

University for the Creative Arts (UCA)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

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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
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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