He6

Propane Combustion Reaction

Chemical Formula: Propane (C₃H₈)
Nature: Colorless, odorless gas used commonly as heating and cooking fuel.
Balanced Equation for Combustion Reaction:
- The reaction of propane with oxygen yields carbon dioxide and water.
- Balanced Equation:
  C₃H₈ + 5 O₂
  ightarrow 3 CO₂ + 4 H₂O

Key Ingredients:
- Ordinary safety matches contain potassium chlorate (KClO₃).
- Potassium chlorate acts as a source of oxygen in combustion reactions.
- Ordinary table sugar (sucrose, C₁₂H₂₂O₁₁) reacts violently with potassium chlorate.
Balanced Equation for Reaction with Sugar:
- Reaction Products: Potassium chloride (KCl), carbon dioxide (CO₂), and water (H₂O).
- Balanced Equation:
  2 KClO₃ + C{12}H{22}O_{11}
  ightarrow 2 KCl + 11 CO₂ + 6 H₂O

Element Representation:
- Element A (represented by red spheres or dark gray) reacts with Element B (represented by blue spheres or light gray).
Balanced Equation for Reaction: A + B ightarrow AB
- Note: Actual symbols for elements A and B depend on the specific elements involved.

Molecular Weight Calculation:
- Sucrose (C₁₂H₂₂O₁₁)
- Molecular Weight Calculation:
  (12 imes 12.01 ext{ g/mol}) + (22 imes 1.01 ext{ g/mol}) + (11 imes 16.00 ext{ g/mol}) = 342.30 ext{ g/mol}
Molar Mass:
- Molar mass of sucrose = 342.30 g/mol.

Weight of Sucrose:
- Given: 2.85 g of sucrose.
Moles Calculation:
- ext{Moles of sucrose} = rac{ ext{mass (g)}}{ ext{molar mass (g/mol)}} = rac{2.85 ext{ g}}{342.30 ext{ g/mol}} ext{ mol}
- Result = 0.00832 mol (rounded to 5 decimal places)

Given: 0.0626 moles of NaHCO₃ (Sodium Bicarbonate)
Grams Calculation:
- Molar mass of NaHCO₃ = 84.01 g/mol
- ext{Grams} = 0.0626 ext{ mol} imes 84.01 ext{ g/mol} = 5.25 ext{ g}

Chemical Reaction Involvement:
- Sodium hypochlorite (NaOCl): known as household bleach.
- Reaction equation:
  2 NaOH(aq) + Cl₂(g)
  ightarrow NaOCl(aq) + NaCl(aq) + H₂O(l)
Grams Calculation of Required NaOH:
- Given: 25.0 g of Cl₂.
- Molar mass of Cl₂ = 70.91 g/mol
- Moles of Cl₂: ext{Moles Cl₂} = rac{25.0 ext{ g}}{70.91 ext{ g/mol}} = 0.352 ext{ mol}
- Stoichiometric ratio: 2 mol NaOH required for 1 mol Cl₂.
- Required moles of NaOH = 2 × 0.352 = 0.704 mol.
- ext{Mass of NaOH} = 0.704 ext{ mol} imes 40.00 ext{ g/mol} = 28.16 ext{ g}

Reaction of Isobutylene:
- Given: 32.8 g of methyl tert-butyl ether obtained from 26.3 g isobutylene reacted with excess methanol.
Balanced Reaction Equation:
C₄H₈(g) + CH₃OH(l)
ightarrow C₅H₁₂O(l)
Theoretical Yield Calculation Needed for Percent Yield:
- Steps needed to obtain theoretical yield to complete percent yield calculation.
- ext{Percent Yield} = rac{ ext{Actual Yield}}{ ext{Theoretical Yield}} imes 100

Given Weight: 40.0 g of ethyl alcohol with a reaction percent yield of 87%.
Balanced Reaction:
2 C₂H₅OH(l)
ightarrow C₄H₁₀O(l) + H₂O(l)
Theoretical Yield Calculation:
- Molar mass of ethyl alcohol = 46.07 g/mol
- ext{Moles of Ethyl Alcohol} = rac{40.0 ext{ g}}{46.07 ext{ g/mol}} ext{ mol}
- Stoichiometry leads us to theoretical yield in grams for diethyl ether and then calculate the actual yield using 87%.

Chemical Reaction for Cisplatin Synthesis:
- Reactants: Potassium tetrachloroplatinate (K₂PtCl₄) and ammonia (NH₃).
Reaction Equation:
K₂PtCl₄(aq) + 2 NH₃(aq)
ightarrow Pt(NH₃)₂Cl₂(s) + 2 KCl(aq)
Given Weights: 10.0 g of K₂PtCl₄ and 10.0 g of NH₃.

Identification of Limiting and Excess Reactants:
- Consider molar masses and amount of moles available for each reactant.
Calculate Moles:
- ext{Moles of K₂PtCl₄} = rac{10.0 ext{ g}}{K(39.10) + 2Pt(195.08) + 4Cl(35.45)}
- Determine the limiting reactant based on stoichiometric needs compared to available moles.
Excess Reactant Calculation:
- Calculate grams consumed and remaining for the limiting reactant.

Product Formation Calculation:
- Moles of product from the limiting reactant leading to grams obtained.
Grams of Cisplatin Calculated:
- Use molecular weights derived from products to calculate using moles formed.

Sulfuric Acid Molarity:
- Given: 2.355 g of H₂SO₄ in 50.0 mL of solution.
- ext{Molarity} = rac{ ext{moles}}{ ext{volume (L)}}
- First, calculate moles of H₂SO₄, then arrive at molarity.

NaOH preparation:
- Desired volume: 500.0 mL, concentration: 0.2500 M from a 1.000 M stock.
- Use dilution equation:
  C₁V₁ = C₂V₂
Solve for V₁, volume needed in mL to achieve desired molarity.

Reaction:
- Hydrochloric Acid (HCl) neutralized by Sodium bicarbonate (NaHCO₃).
Equation:
HCl(aq) + NaHCO₃(aq)
ightarrow NaCl(aq) + H₂O(l) + CO₂(g)
Calculation of Volume Needed:
- Given: 18.0 mL of 0.100 M HCl.
- Use stoichiometry to determine volume of 0.125 M NaHCO₃ required to neutralize the acid.