Chemistry Answer Note #29

Question 1 What amount of chlorine gas produces 2.24mol of phosphorus trichloride? Answer In order to determine the amount of chlorine gas that produces 2.24mol of phosphorus trichloride, we must first understand the reaction between these two substances. The reaction between chlorine gas and phosphorus trichloride is as follows: PCl3 + Cl2 → PC l5. According to the balanced equation, for every 1 mol of PCl3 produced, 1 mol of Cl2 is consumed. Therefore, to produce 2.24 mol of PCl3, we would need 2.24 mol of Cl2. Calculation method: Step 1: Write out the balanced equation for the reaction: PCl3 + Cl2 → PC l5. Step 2: Determine the mole ratio between PCl3 and Cl2. According to the balanced equation, the mole ratio is 1:1. Step 3: Set up a proportion to determine the amount of Cl2 needed to produce 2.24 mol of PCl3: 1 mol of PCl3 / 1 mol of Cl2 = 2.24 mol of PCl3 / x mol of Cl2. Step 4: Solve for x: x = (2.24 mol of PCl3 * 1 mol of Cl2) / 1 mol of PCl3 = 2.24 mol of Cl2. Therefore, in order to produce 2.24 mol of PCl3, we would need 2.24 mol of Cl2. Note: It is always important to check if the units cancel out and the answer makes sense in terms of the given information. In this case, the units of mol of PCl3 cancel out and we are left with mol of Cl2, which is the unit we are looking for. The answer also makes sense since we are producing more PCl3 than Cl2, indicating that Cl2 is the limiting reactant. Question 2 Nitric acid can be produced by reacting nitrogen gas, oxygen gas and water according to the equation: 2N2(g) + 5O2(g) + 2H2O(l) --> 4HNO3(aq). If 5.85g of nitrogen gas and 14.05g of oxygen gas react with an excess of water, how many grams of HNO3 will form? Answer According to the balanced chemical equation, 2 moles of nitrogen gas (N2) react with 5 moles of oxygen gas (O2) and 2 moles of water (H2O) to produce 4 moles of nitric acid (HNO3). In this scenario, we are given 5.85g of nitrogen gas and 14.05g of oxygen gas. In order to determine the mass of HNO3 formed from the given amounts of reactants, we first need to convert the given masses into moles. The molar mass of N2 is 28 g/mol and the molar mass of O2 is 32 g/mol. Therefore, 5.85g of N2 is equivalent to 5.85g/28 g/mol = 0.2089 mol of N2. Similarly, 14.05g of O2 is equivalent to 14.05g/32 g/mol = 0.4391 mol of O2. Next, we need to determine the limiting reagent in the reaction. This is the reactant that will be completely used up in the reaction and will limit the amount of product that can be formed. To do this, we can use the mole ratio between N2 and O2 in the balanced equation. The mole ratio between N2 and O2 is 2:5. This means that for every 2 moles of N2, we need 5 moles of O2 to react completely. However, in this case, we have 0.2089 mol of N2 and 0.4391 mol of O2. This shows that there is an excess of O2 and N2 is the limiting reagent. Using the mole ratio between N2 and HNO3, we can determine the theoretical yield of nitric acid that can be formed from 0.2089 mol of N2. The mole ratio between N2 and HNO3 is 2:4, which means that for every 2 moles of N2, we can produce 4 moles of HNO3. Therefore, the theoretical yield of HNO3 from 0.2089 mol of N2 is 4 x 0.2089 mol = 0.8356 mol. Finally, we need to convert the theoretical yield of HNO3 from moles to grams. The molar mass of HNO3 is 63 g/mol. Therefore, the mass of HNO3 that can be formed from 0.8356 mol is 0.8356 mol x 63 g/mol = 52.6368 g of HNO3. Question 3 Aluminum reacts with bromine to form aluminum bromide in the reaction: 2Al(s) + 3Br2(l) --> 2AlBr3. If 15.79g of aluminum react with 55.61g of bromine, how many grams of aluminum bromide will form? Answer To accurately determine the amount of aluminum bromide formed in this reaction, we need to use stoichiometric calculations. Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. First, let's write a balanced chemical equation for the given reaction: 2Al(s) + 3Br2(l) --> 2AlBr3 According to the law of conservation of mass, the total mass of the reactants must equal the total mass of the products. Therefore, the molar ratio of aluminum to aluminum bromide is 2:2, or 1:1. This means that for every 2 moles of aluminum, we will get 2 moles of aluminum bromide. Next, we need to convert the given masses of aluminum and bromine into moles. The molar mass of aluminum is 26.98 g/mol and the molar mass of bromine is 79.904 g/mol. Number of moles of aluminum: 15.79g / 26.98 g/mol = 0.585 moles Number of moles of bromine: 55.61g / 79.904 g/mol = 0.695 moles Based on the balanced chemical equation, we can see that a 1:1 molar ratio of aluminum to aluminum bromide is needed. Since we have 0.585 moles of aluminum, we will also need 0.585 moles of aluminum bromide. Now, we can convert the moles of aluminum bromide back into grams: Mass of aluminum bromide = 0.585 moles x 266.70 g/mol (molar mass of AlBr3) = 156.06 g Therefore, 156.06 grams of aluminum bromide will be formed in this reaction. Question 4 What are the balanced coefficients of the reaction CaO(s) + H2O(l) --> Ca(OH)2(aq)? Answer To answer this question, we first need to understand the basics of chemical reactions and how to balance them. A balanced chemical equation is one that has the same number of atoms of each element on both sides of the equation. This is achieved by adjusting the coefficients, which are the numbers in front of the reactants and products. In the given reaction, we have CaO (calcium oxide) and H2O (water) as the reactants, and Ca(OH)2 (calcium hydroxide) as the product. To balance this equation, we need to ensure that the number of calcium atoms, oxygen atoms, hydrogen atoms, and hydroxide ions (OH^-) are the same on both sides. Step 1: Write the unbalanced equation CaO + H2O --> Ca(OH)2 Step 2: Count the number of each type of atom on both sides of the equation. Reactants: Ca - 1 atom, O - 1 atom, H - 2 atoms Products: Ca - 1 atom, O - 2 atoms, H - 2 atoms Since the number of oxygen atoms is not balanced, we will start by balancing it first. Step 3: Add a coefficient in front of Ca(OH)2 to balance the number of oxygen atoms. In this case, we need 2 molecules of Ca(OH)2 to have a total of 2 calcium atoms and 4 oxygen atoms on the product side. CaO + H2O --> 2Ca(OH)2 Step 4: Now, let's check the number of hydrogen atoms. On the product side, we already have 4 hydrogen atoms. So, we need to add a coefficient of 2 in front of H2O to have a total of 2 hydrogen atoms on the reactant side as well. CaO + 2H2O --> 2Ca(OH)2 Step 5: Finally, let's check the number of calcium atoms. On the reactant side, we have 1 calcium atom, and on the product side, we have 2 calcium atoms. To balance this, we just need to add a coefficient of 2 in front of CaO on the reactant side. 2CaO + 2H2O --> 2Ca(OH)2 This means that the balanced coefficients for the given reaction are 2 for CaO, 2 for H2O, and 2 for Ca(OH)2. Question 5 How many molecules of magnesium oxide will form when excess magnesium reacts with 480 molecules of oxygen Answer According to the balanced chemical equation for the reaction between magnesium and oxygen, 2Mg + O2 → 2MgO, we can see that for every 2 molecules of magnesium that react, we will also need 1 molecule of oxygen to form 2 molecules of magnesium oxide. Therefore, if we have 480 molecules of oxygen, we will need twice as many molecules of magnesium, which is equal to 960 molecules, to fully react with all of the oxygen. This means that 960 molecules of magnesium oxide will form when excess magnesium reacts with 480 molecules of oxygen.