unit 2 - chemistry (1)

ThiChung Le Unit 2 assignment: Chemical reactions and stoichiometric relationships 1. a. 2KCLO 3 → 2KCL + 3O 2 b. 2ALBr 3 + 3K 2 SO 4 → 6KBr + Al 2 (SO 4 ) 3 c. FeCL 3 + 3NAOH → Fe(OH) 3 + 3NACl 2. Write and balance the skeleton equation for each of the following chemical reactions: [6 marks] a. Potassium metal and chlorine gas combine to form potassium chloride. 2K + Cl 2 → 2KCl b. Sodium phosphate and calcium chloride react to form calcium phosphate and sodium chloride. 2Na 3 PO 4 + 3CaCl 2 → Ca 3 (PO 4 ) 2 + 6NaCl c. Zinc and lead(II) nitrate react to form zinc nitrate and lead. Zn + Pb(NO 3 ) 2 → Zn(NO 3 ) 2 + Pb 3. Identify each of the following reactions as synthesis, decomposition, single displacement, or double displacement: [6 marks] a. Sodium + chlorine → sodium chloride synthesis b. 3Ca(OH) 2 +Al 2 (SO4) 3 → 3CaSO 4 + 2Al(OH) 3 double displacement c. 2PbSO 4 → 2PbSO 3 + O 2 decomposition d. Ammonium chloride + mercury( I ) acetate → ammonium acetate + mercury ( I ) chloride double displacement e. 3Mg + Fe 2 O 3 → 2Fe + 3MgO single displacement

f. 2NH 3 + H 2 SO 4 → (NH 4 )2SO 4 synthesis 4. Complete each of the following neutralization reactions. Word equations should be completed as word equations, and chemical equations should be completed as balanced chemical reactions. [4 marks] a. H 2 SO 4 + 2NH 4 OH → (NH 4 ) 2 SO4 + 2H2O b. Aluminium hydroxide + Sulfuric acid → Sluminium sulphate and Water 5. Complete and balance each of the following combustion reactions for pentane (C 5 H 12 ), one of the fuels in gasoline: [4 marks] a. Complete combustion C 5 H 12 + 8O 2 → 5CO 2 + 6H 2 O b. Incomplete combustion C 5 H 12 + 6O 2 → 4CO + CO2 +6H2O 6. Type of reaction Reactants products Potential qualitative evidence synthesis 2Mg (s) + O2 (g) 2MgO(s) Magnesium oxide is a white solid. you observe a white powder or sol being formed during the reaction. single displacement Zn (s) + 2HCl (aq) ZnCl 2 (aq)+ H 2 (g) Bubbles may appear as gas being released from the solution. double displacement NaCl (aq) + AgNO 3(aq) AgCl(s)+NaNO 3 (a q) white precipitate of silver chloride may form. The appearance of a white, cloudy or milky substance indicates the presence of the precipitate. combustion CH4 (g) + 2O 2(g) CO 2 (g)+ 2H 2 O(g) The appearance of a flame during the reaction

In the presence of an electrolyte, such as water or an acidic solution, zinc ions (Zn 2+ ) are formed through the oxidation of zinc atoms on the surface of the galvanized part. 9. a.I n what way is the combustion of sulfur harmful to the environment? [1 mark] The combustion of sulfur can be harmful to the environment primarily due to the formation of sulfur dioxide (SO2) gas. Sulfur dioxide is a major air pollutant that contributes to various environmental issues. It can lead to the following harmful effects. Air pollution: Sulfur dioxide is a primary contributor to air pollution. When released into the atmosphere during the combustion of sulfur-containing fuels, such as coal or petroleum, it can react with other pollutants and sunlight to form fine particulate matter (PM2.5) and smog. These pollutants can reduce air quality and contribute to respiratory problems, especially for individuals with pre-existing conditions such as asthma or other respiratory illnesses. Acid rain: Sulfur dioxide emissions can contribute to the formation of acid rain. When sulfur dioxide reacts with water vapor and other atmospheric compounds, it forms sulfuric acid. Acid rain can have detrimental effects on ecosystems, including the acidification of lakes, rivers, and soil. It can harm aquatic life, damage vegetation, and erode buildings and infrastructure. Health impacts: Inhalation of sulfur dioxide can have negative health effects on humans and animals. It can irritate the respiratory system, leading to respiratory symptoms such as coughing, wheezing, and shortness of breath. Prolonged exposure to high levels of sulfur dioxide can aggravate existing respiratory conditions, increase the risk of respiratory infections, and contribute to the development of cardiovascular diseases. b. SO 2 is very harmful to the environment as they react with water molecules H 2 O during rain to form sulfurous acid( H 2 SO 3 ) and also when oxidized to sulfur trioxide( SO 3 ) reacts with water molecules to form sulphuric acid (H 2 SO 4 ) and falls on earth's surface as acid rain. The reaction involved: SO 2 + H 2 O ⟶ H 2 SO 3 2SO 2 + O 2 ⟶ 2SO 3 H 2 O+ SO 3 ⟶ H 2 SO 4

10. In my opinion, labelling the oil as “ dirty” is acceptable. It brings a huge negative effect to our planet. Burning of fossil fuels like oil, coal and gas is driving one of the biggest challenges facing the world today: climate change. Extreme weather events, rising oceans, and record setting temperatures are wreaking havoc on hundreds of millions of lives and livelihoods around the world. Greenhouse gas emissions, primarily from the burning of fossil fuels, have already warmed the globe by more than 1°C since the beginning of the industrial revolution. Unless we can rein in these emissions and ambitiously transition to a just, clean, and renewable energy future - the planet will become unrecognizable as global temperatures soar by 4, 5, or 6 °C and beyond. Many environmental activist groups continue to use the term "dirty oil" when referring to oil sands production. These groups contend the oil sands contribute disproportionately to global warming, use vast quantities of fresh water, release contaminated water from leaky tailings ponds, and ignore local Indigenous groups. Such statements undermine the tremendous work the oil sands industry has been undertaking for more than a decade to improve environmental, social and governance performance, and reduce impacts. Moverover, The vast majority of the historical global emissions that are driving climate change have come from developed countries, like the United States, the EU, Japan, and Canada; but it is the poorest countries - those who can least afford to adapt to a changing climate - that are suffering first and worst. The recent global climate agreement in Paris was a major step in recognizing the global urgency of the crisis, but it will take serious action from national and subnational governments to meet new goals that aspire to limit global warming to 1.5°C. The deal also fell short on critical dimensions of equity, climate financing, and efforts to support the most vulnerable when even adaptation becomes impossible. Over the past decade, Todd Crawford from the Conference Board has estimated that refining has consistently returned an average annual profit margin of 11%, making it the least lucrative sector in the energy supply chain. Unlike the global crude market, refining is subject to the demands of local markets. In the province of Alberta, which is abundant in oil, the oil refining sector only contributes a small 0.3 percent to the GDP. At a meeting with Canadian directors, it was concluded that constructing a new refinery would come with a hefty $7 billion price tag. Trying to sell an investment that yields diminishing returns is challenging, especially considering that there are around 60 mostly inactive refineries along the Gulf Coast of the UnitedStates. Despite their defeat in negotiations with EU policymakers, certain environmentalists claim victory in a broader dispute. Canadian oil is filthy from the Europeans perspective for the right reasons.

n CO2 = n C6H14 x ( y mol CO 2 / z mol C 6 H 14 ) n CO2 = 0.371 mol C 6 H 14 x (6 mol CO 2 /1 mol C 6 H 14 ) n CO2 = 2.226 mol CO 2 Calculate mass of carbon dioxide is produced: m CO2 = n CO2 x M CO2 = 2.226 mol x ( 12.001 + 2 x 15.999) g/mol = 97.94 g Therefore, the mass of carbon dioxide produced by the complete combustion of 32 g of hexane is 97.94 g. 13. Use mole ratios to identify the limiting reagent in the following reaction: 2.10 mol of sodium mixes with 1.05 mol of water to form sodium hydroxide and hydrogen gas. [2 marks] Balanced chemical equation: 2Na + 2 H 2 O → 2 NaOH + H 2 From the equation, we can see that 2 moles of sodium react with 2 moles of water to produce 2 moles of sodium hydroxide and 1 mole of hydrogen gas. We have: n Na : 2.1 mol n H20 : 1.05 mol To determined the limiting reagent: First, we compare the mole ratio of sodium to sodium hydroxide: n NaOH = n Na x ( y mol NaOH / z mol Na ) n NaOH = 2.10 mol Na x ( 2 mol NaOH / 2 mol Na ) n NaOH = 2.10 mol NaOH Second, we compare the mole ratio of water to sodium hydroxide: n NaOH = n H2O x ( y mol NaOH / z mol H2O ) n NaOH = 1.05 mol H2O x ( 2 mol NaOH / 2 mol H2O) n NaOH = 1.05 mol NaOH From the calculations, we can see that both reactants will produce the same amount of sodium hydroxide (NaOH). However, the stoichiometric ratio tells us that 2 moles of

sodium react with 2 moles of water to produce 2 moles of sodium hydroxide. Therefore, water is the limiting reagent in this reaction since it is present in a smaller amount than the stoichiometric ratio requires. So water (H2O) is the limiting reagent in this reaction. 14. Determine the number of excess moles of the excess reagent in the following reaction: 2.40 mol of magnesium mixing with 1.65 mol of oxygen gas, making magnesium oxide. [2 marks] Write a balanced chemical reaction: 2Mg + O2 → 2MgO We have: n Mg = 2.40 mol, n O2 = 1.65 mol To determine the excess reagent, Assume that all of the magnesium reacts. Find the number of O2 ( mol ) n O2 = nMg x ( y mol O 2 / z mol Mg ) n O2 = 2.40 mol Mg x ( 1 mol O 2 / 2 mol Mg ) n O2 = 1.2 mol O 2 So the amount of 2.4 mol Mg require 1.2 mol O 2 to react completely. But there are 1.65 mol O 2 . There are , O 2 is excess reagent. To determine the number of excess moles of the excess reagent: Excess moles of O 2 = 1.65 mol - 1.2 mol = 0.45 mol Therefore, there are 0.45 moles of excess O 2 in the reaction. 15. 1. Distinguish between the terms “actual yield” and “theoretical yield.” [2 marks] Theoretical yield: The theoretical yield refers to the maximum amount of product that can be obtained from a chemical reaction under ideal conditions. It is calculated based on the stoichiometry of the balanced chemical equation and assuming that the reaction proceeds to completion without any side reactions or losses. The theoretical yield is often expressed in moles or grams

We have: m HNO3 = 145g, m NH3 = 55 g Solve for the number of moles: n HNO3 = m HNO3 / M HNO3 n HNO3 = m HNO3 / ( MH + MN+ 3 Х MO) n HNO3 = 145 g / ( 1.007+ 14.007 + 3 Х 15.999) g/mol n HNO3 = 2.3 mol n NH3 = m NH3 /M NH3 n NH3 = mNH3 / ( MN + 3 Х MH) n NH3 = 55 g / ( 14.007 + 3 Х 1.007)g/mol n NH3 = 3.23 mol Since the stoichiometric ratio between hydrogen nitrate and ammonium nitrate is 1:1, the limiting reagent will be the reactant that has a smaller number of moles. In this case, hydrogen nitrate has 2.30 moles and ammonia has 3.23 moles. So hydrogen nitrate is the limiting reagent. So n NH4NO3 = n HNO3 Х ( y mol NH 4 NO 3 / z mol HNO 3 ) n NH4NO3 = 2.3 mol HNO 3 Х ( 1 mol NH 4 NO 3 / 1 mol HNO 3 ) n NH4NO3 = 2.3 mol We have: mNH4NO3 = n NH4NO3 Х MHN 4 NO 3 = 2.3 mol Х 80.043g = 184.09 g So, approximately 184.09 grams of ammonium nitrate can be produced from the given amounts of hydrogen nitrate and ammonia. b. If the actual yield is only 181 g, what is the percent yield? [3 marks] We have the theoretical yield of ammonium nitrate from the given amounts of hydrogen nitrate and ammonia is approximately 184.09 g Percentage yield = ( actual yield / theoretical yield ) Х 100%

Percentage yield = ( 181 g / 184.09 g ) Х 100% Percentage yield = 98.32% Therefore, the percent yield is 98.32% 17. a. 2C 2 H 3 NO + H 2 0 → C 3 H 8 N 2 O + CO 2 We have m C3H8N2O = 42 000 kg M C2H3NO = (12Х2)+(1Х3)+14+16 = 57 g M C3H8N2O = (12Х3)+(1Х8)+(14Х2)+16 = 88g M CO2 = 12+(16Х2) = 44g First on this question, we make use of mole to mole comparison 2 mole of C 2 H 3 NO yields 1 mole of C 3 H 8 N 2 O For that, we calculate moles of C 2 H 3 NO = Given mass / Molar mass n C2H3NO = (42000 Х10 3 g ) / 57 g n C2H3NO = 736.8Х10 3 g We have: n C 3 H 8 N 2 O= n C2H3NO Х ( y mol C 3 H 8 N 2 O / z mol C 2 H 3 NO ) n C 3 H 8 N 2 O = 736.8Х10 3 g Х ( 1 mol C 3 H 8 N 2 O / 2 mol C 2 H 3 NO ) n C 3 H 8 N 2 O = 368.4 Х 10 3 mol C 3 H 8 N 2 O So m C3H8N2O = n C3H8N2O Х M C3H8N2O = ( 368.4 Х 10 3 mol ) Х 88 m C3H8N2O = 32419.2 Х 10 3 g We have nCO 2 = n C2H3NO Х ( y mol CO 2 / z mol C 2 H 3 NO ) nCO 2 = 736.8Х10 3 g Х (1 mol CO 2 / 2 mol C 2 H 3 NO ) nCO 2 = 368.4 Х 10 3 mol mCO 2 = nCO 2 Х MCO 2 = (368.4 Х 10 3 mol ) Х 44g = 16209 Х 10 3 g So to have 42 000 kg (42 tonnes) of methyl isocyanate reacted with excess water, we need to use CO 2 is 16209 Х 10 3 g and 3 H 8 N 2 O is 32419.2 Х 10 3 g. b. List of Mistakes

Process: Materials Used: ● Solid sodium carbonate (Na2CO3) ● 6 M hydrochloric acid (HCl) ● Electronic balance ● Evaporating dish ● Watch glass (suitable for covering the evaporating dish) ● Stand and ring clamp ● Wire gauze ● Dropper pipette ● Stirring rod ● Bunsen burner Procedure: 1. Begin by determining and recording the mass of a clean and dry evaporatingdish. 2. Measure an approximate quantity of 0.5grams of solid sodium carbonate Na2CO3 and record the mass. 3. Carefully add this quantity to the evaporating dish, and record the collective mass of the evaporating dish, the watch glass, and the sodium carbonate (Na2CO3). 4. Using the dropper pipette, slowly add the hydrochloric acid (HCl) drop by drop into the sodium carbonate (Na2CO3) present in the evaporating dish. Use approximately 5mL of HCl 5. Observe the bubbling phenomenon. Continue the addition of HCl until the bubbling ceases entirely, indicating the completion of the reaction where all Na2CO3 has reacted. 6. Assemble the stand, ring clamp, and wire gauze apparatus for heating. Cover the evaporating dish with the watch glass and position it on the wire gauze. 7. Apply gentle heat to the solution using the Bunsen burner flame. This is useful for eliminating the H2O produced during the reaction and any surplus HCl that might be present. 8. Continue heating until the contents in the evaporating dish are entirely dry. An additional two minutes of heating is recommended after the initial perception of dryness, ceasing if the white salt starts to turn brownish. 9. Allow the evaporating dish to cool down to room temperature before proceeding to the next step.

10.Measure and record the mass of the evaporating dish residue, primarily composed of sodium chloride ( NaCl ). 11. Properly dispose of the waste generated from this experiment in the sink. Explanation: Lets write the equation for the reaction The reaction for the experiment can be written as : Na 2 CO 3 (s)+2HCl (aq) ⟶ 2NaCl (aq )+CO 2 (g)+H2O(l) s=solid aq=aqueous l=liqid This equation signifies the formation of aqueous sodium chloride, carbon dioxide gas, and liquid water upon the completion of the double displacement reaction between sodium carbonate and hydrochloric acid.