Chemistry Answer Note #38

Question 1 Oxygen and carbon disulfide react to form carbon dioxide and sulfur dioxide, like this: 302(g)+CS2(g)-CO2(g)+2S02(g) Use this chemical equation to answer the questions in the table below. Suppose 210.mmol of O2 and 70.0 mmol of CS2 are added to an empty flask. How much CO2 will be in the flask at equilibrium? Answer To answer this question, we must first understand the chemical equation and its corresponding balanced coefficients. From the given equation, we know that 3 moles of oxygen gas (O2) react with 1 mole of carbon disulfide (CS2) to form 1 mole of carbon dioxide (CO2) and 2 moles of sulfur dioxide (SO2). This means that for every 3 molecules of O2 and 1 molecule of CS2 added to the flask, 1 molecule of CO2 and 2 molecules of SO2 will be produced at equilibrium. With the given information, we first need to convert the amounts of O2 and CS2 from millimoles (mmol) to moles. One mole is equal to 1000 mmol, so we have 0.210 mol of O2 (210 mmol * 1 mol/1000 mmol) and 0.070 mol of CS2 (70.0 mmol * 1 mol/1000 mmol). Next, we need to determine the limiting reactant in this reaction. The limiting reactant is the substance that is completely used up in the reaction, and therefore determines the maximum amount of product that can be formed. To find the limiting reactant, we need to calculate the number of moles of each reactant that are required for the reaction to take place. From the balanced equation, we can see that 3 moles of O2 are needed for every 1 mole of CS2. Therefore, we need 3 times more moles of O2 than CS2 for the reaction to proceed. In this case, we have 0.070 mol of CS2 which is less than 3 times the amount of O2 (0.210 mol). This means that CS2 is the limiting reactant and will be completely consumed in the reaction. Therefore, only 0.070 mol of CO2 will be produced at equilibrium. Now, to answer the question, we simply need to convert the moles of CO2 to millimoles (mmol) by multiplying 0.070 mol by 1000 mmol/1 mol. This gives us 70 mmol of CO2 in the flask at equilibrium. In summary, in an empty flask when 210 mmol of O2 and 70.0 mmol of CS2 are added, the amount of CO2 at equilibrium will be 70 mmol. Question 2 If you mix 50.0 grams of baking soda with 100.0 grams of vinegar, how many grams of carbon dioxide gas will be produced? Assume that the reaction goes to completion and there is no excess of either reactant. Answer To calculate the amount of carbon dioxide gas produced in this reaction, we must first determine the balanced chemical equation for the reaction between baking soda (sodium bicarbonate, NaHCO3) and vinegar (acetic acid, CH3COOH). The balanced equation is: NaHCO3 + CH3COOH → NaCH3COO + H2O + CO2 From this equation, we can see that for every 1 mole of NaHCO3 reacted, 1 mole of CO2 will be produced. Therefore, we need to convert the given amounts of baking soda and vinegar into moles to accurately calculate the amount of CO2 produced. To convert grams to moles, we will use the molar mass of each substance. The molar mass of NaHCO3 is 84.0066 g/mol and the molar mass of CH3COOH is 60.05 g/mol. Calculation for moles of NaHCO3: 50.0 g NaHCO3 x (1 mol NaHCO3/84.0066 g NaHCO3) = 0.595 mol NaHCO3 Calculation for moles of CH3COOH: 100.0 g CH3COOH x (1 mol CH3COOH/60.05 g CH3COOH) = 1.664 mol CH3COOH According to the balanced equation, 1 mole of NaHCO3 reacts with 1 mole of CH3COOH to produce 1 mole of CO2. Thus, we can conclude that 0.595 mol of CO2 will be produced in this reaction. To convert moles back to grams, we simply multiply by the molar mass of CO2, which is 44.01 g/mol. Calculation for grams of CO2: 0.595 mol CO2 x 44.01 g/mol = 26.186 g CO2 Therefore, mixing 50.0 grams of baking soda and 100.0 grams of vinegar will produce 26.186 grams of carbon dioxide gas. Question 3 Why is burning of candle both a physical and chemical change? Answer Burning of a candle is an everyday event, yet it involves both physical and chemical changes. A physical change is a change in the physical properties of a substance, while a chemical change is a change in the chemical composition of a substance. In order to understand why the burning of a candle is both a physical and chemical change, we must first examine the components and reactions of a candle. A candle is made up of three main components: wax, wick and a flame. The wax is usually paraffin, which is a type of hydrocarbon. The wick is made of a fibrous material, such as cotton, that is able to absorb and draw up the liquid wax. When a candle is first lit, the wax at the base of the wick is vaporized by the heat of the flame. This vapor then travels up the wick and reacts with the oxygen in the air, causing the wax vapor to combust, or burn. The burning of the candle involves both physical and chemical changes. The physical change can be seen in the melting of the solid wax into liquid. This is a change in physical state, from solid to liquid, but the chemical composition of the wax remains the same. The liquid wax is then drawn up the wick through capillary action, another physical process. As the wax vapor rises with the heat of the flame, it undergoes a chemical change. The hydrocarbons in the wax combine with oxygen to produce carbon dioxide, water vapor, and heat energy. This is evident by the presence of a flame, which is the result of a chemical reaction between the wax and oxygen. So why is burning of candle both a physical and chemical change? It is because while the wax undergoes a physical change in its state, the chemical composition of the wax is also changing due to the chemical reaction with oxygen. These two changes occur simultaneously and are necessary for a candle to burn. Question 4 Easy way to memorize the colloidal table of dispersion phase and medium Answer As a high achiever in my high school chemistry class, I have found that the best way to memorize the colloidal table of dispersion phase and medium is through practice and visual aids. The colloidal table categorizes dispersions into four main types: sol, gel, emulsion, and foam; based on their dispersion medium (solid, liquid, or gas) and their dispersed phase (molecules, particles, or droplets). To start, it is important to understand the definitions of dispersion and dispersion medium. Dispersion refers to the process of dispersing small particles or droplets throughout a medium, while the dispersion medium is the substance in which the particles or droplets are dispersed. Keeping this in mind, let's take a look at how to memorize the colloidal table. Firstly, make a simplified table or chart with the four types of dispersions on one side and their corresponding dispersion medium and dispersed phase on the other side. For example, under "Sol" you can write "solid dispersed in liquid" and under "Gel" you can write "solid or liquid dispersed in solid." This visual aid will help you quickly recall the different types of dispersions and their components. Next, practice categorizing common examples of colloids into the correct type of dispersion. For example, mayonnaise would go under "emulsion" as it is a liquid dispersed in another liquid, while whipped cream would fall under "foam" with gas dispersed in a liquid. The more you practice, the easier it will become to identify the correct type of dispersion. Additionally, it may be helpful to create mnemonics or acronyms to remember the different types of dispersions. For example, "SOGeF" can stand for "sol, gel, emulsion, foam" to help you remember the order of the four types. Question 5 by adding NaCl to a saturated aqueous soltuion of PbCl2 what happens? Answer The addition of NaCl to a saturated aqueous solution of PbCl2 results in the formation of a precipitate. This reaction is known as a double displacement reaction, where the cations of one compound switch places with the cations of another compound, resulting in the formation of a new compound. When NaCl is added to a saturated solution of PbCl2, the following reaction takes place: PbCl2 (aq) + 2NaCl (aq) → PbCl2 (s) + 2NaCl (aq) This reaction can be broken down into two steps. In the first step, the Pb2+ ions from the PbCl2 solution react with the Cl- ions from the NaCl solution to form PbCl2, which is a solid compound. This is because the Pb2+ ions have a stronger affinity for Cl- ions compared to the Na+ ions. This reaction is known as a precipitation reaction. In the second step, the resulting solution is still saturated in NaCl, meaning that there is a high concentration of Na+ and Cl- ions in the solution. This leads to the formation of more PbCl2, resulting in a net increase in the amount of precipitate formed.