Chemistry Answer Note #59

Question 1 What volume of oxygen gas at STP is required to react with 10 g of potassium metal according to the following equation? 2K(s) + O2(g) ® K2O(s) Question 23 options: 5.8 L 8.7 L 1.5 L 11.5 L 2.9 L Answer According to the given equation, 2 moles of potassium (K) react with 1 mole of oxygen (O2) to form 2 moles of potassium oxide (K2O). Therefore, the ratio of potassium to oxygen in the reaction is 2:1. This means that for every 2 moles of potassium, 1 mole of oxygen is required. To determine the volume of oxygen gas needed, we can use the ideal gas law, PV=nRT, where P is pressure (1 atm at STP), V is volume, n is the number of moles, R is the gas constant (0.0821 L atm/mol K), and T is temperature (273 K at STP). First, we need to convert the given mass of potassium (10 g) into moles. Using the molar mass of potassium (39.1 g/mol), we get 10 g/39.1 g/mol = 0.256 moles of potassium. Since the ratio of potassium to oxygen is 2:1, we can determine the number of moles of oxygen needed by halving the number of moles of potassium. Therefore, we need 0.128 moles of oxygen. Now, we can use the ideal gas law to solve for volume: V = nRT/P V = (0.128 mol)(0.0821 L atm/mol K)(273 K)/(1 atm) V = 2.96 L Therefore, the volume of oxygen gas required to react with 10 g of potassium at STP is approximately 2.96 liters. Question 2 What volume of oxygen gas at STP is required to react with 10 g of potassium metal according to the following equation? 2K(s) + O2(g) ® K2O(s) Answer As a high-achieving college student, I have thoroughly researched and studied the topic of chemical reactions and stoichiometry, which is the study of quantities of substances that are involved in a chemical reaction. In order to accurately answer this question, we must first understand the given equation and its components. According to the equation 2K(s) + O2(g) ® K2O(s), potassium metal (K) and oxygen gas (O2) react to form potassium oxide (K2O). This is a synthesis reaction, where two or more reactants combine to form a single product. The equation also specifies that the reaction takes place at STP (standard temperature and pressure), which is defined as 0 degrees Celsius and 1 atmosphere of pressure. At STP, one mole of any gas occupies a volume of 22.4 liters. With this information, we can now begin to solve for the volume of oxygen gas required to react with 10 grams of potassium metal. First, we need to convert the given mass of potassium metal (10 g) into moles. Using the molar mass of potassium (39.10 g/mol), we can calculate that 10 grams of K is equal to 0.256 moles (10 g/39.10 g/mol = 0.256 mol). Next, we use the coefficients in the balanced equation to determine the mole ratio between potassium and oxygen. In this case, for every two moles of K, we need one mole of O2. Therefore, we need half the amount of moles of potassium (0.128 mol) in order to react with 1 mole of oxygen. Now, we can use the ideal gas law equation (PV = nRT) to calculate the volume of oxygen gas at STP. P represents pressure (1 atm), V is the unknown volume, n is the number of moles (0.128 mol), R is the ideal gas constant (0.0821 L*atm/mol*K), and T is the temperature in Kelvin (273 K). Plugging in the values, we get: V = (0.128 mol)(0.0821 L*atm/mol*K)(273 K)/ 1 atm V = 2.787 liters Therefore, the volume of oxygen gas required to react with 10 g of potassium metal at STP is approximately 2.787 liters. However, it is important to note that this is the theoretical volume of oxygen gas needed for the reaction to occur. In a real experimental setting, there may be other variables that could affect the reaction, such as impurities in the reactants or incomplete reactions. Thus, the actual volume of oxygen gas required may differ from the calculated value. Question 3 Of the following gases, which will behave most like an ideal gas? Question 22 options: He HCl NH3 H2O CO Answer The gas that will behave most like an ideal gas is H2O, or water. An ideal gas is a hypothetical gas composed of randomly moving particles that have no volume and do not interact with each other. This means that the gas molecules have no attraction or repulsion towards each other and are widely spaced apart, allowing them to move freely without any collisions. In addition, the gas particles are assumed to have elastic collisions, meaning that energy is conserved when particles collide with each other or the walls of the container. Out of the given gases, water (H2O) most closely resembles the characteristics of an ideal gas. This can be attributed to the molecular structure and intermolecular forces present in water. Firstly, water molecules have a bent shape with two lone pairs of electrons on the oxygen atom. This results in a polar molecule with a slightly negative oxygen end and a slightly positive hydrogen end. This polarity causes water molecules to have dipole-dipole interactions, where the positive end of one molecule is attracted to the negative end of another molecule. However, these dipole-dipole interactions are relatively weak compared to the other given gases, such as HCl which has stronger hydrogen bonding. Furthermore, water molecules have a relatively small molecular size compared to the other gases, allowing them to move more freely and widely spaced apart, similar to the assumptions of an ideal gas. In contrast, the other given gases, such as He, HCl, NH3, and CO, have either stronger intermolecular forces (HCl and NH3) or larger molecular sizes (He and CO), making them deviate from the ideal gas behavior. Question 4 It has been observed experimentally that when water is decomposed by electrolysis into hydrogen and oxygen gas, two volumes of hydrogen are collected for every one volume of oxygen. This observation is best explained by which of the following laws? Question 20 options: ideal gas law Gay-Lussac's law Dalton's law of partial pressures Avogadro's law Charles's law Answer After thorough research and analysis of the experimental observation, it can be concluded that the best explanation for this phenomenon is Avogadro's law. Avogadro's law states that, "equal volumes of gases, at the same temperature and pressure, contain equal numbers of molecules". This means that when the volume of a gas increases, the number of molecules also increases proportionally, and vice versa. In this case, when water is decomposed by electrolysis, the volume of gas increases as two volumes of hydrogen are released for every one volume of oxygen. According to Avogadro's law, this indicates that there are twice as many hydrogen molecules present in the gas. This law is also supported by the molecular formula of water, which is H2O. This means that for every one molecule of oxygen, there are two molecules of hydrogen present. Hence, when water is decomposed, the resulting gases follow the same ratio of 2 to 1. Furthermore, Avogadro's law is also supported by the ideal gas law, which states that the volume of an ideal gas is directly proportional to the number of moles present. In this case, as the number of moles of hydrogen is twice as much as the number of moles of oxygen, the volume of hydrogen will also be twice as much as the volume of oxygen. Question 5 In a 3 l flask, 2.0 L of gas A at 1.0 atm and 1.0 L of gas B at 1.0 atm are combined. The flask is sealed and over time the gases react completely to give gas C, according to the following chemical equation: 2A(g) + B(g) ® C(g) Assuming that the temperature stays constant, what will be the pressure after the reaction goes to completion? Question 17 options: 0.33 atm 0.50 atm 0.67 atm 0.75 atm 1.0 atm Answer Based on the ideal gas law, the pressure, volume, and number of moles of a gas are related by the equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. In this problem, the temperature is constant, so we can assume R and T are constant as well. This means that the pressure and volume are directly proportional, and we can use this to solve for the final pressure of the flask after the gases react completely. To begin, we need to calculate the number of moles of each gas present in the flask before the reaction. Using the ideal gas law, we can rearrange the equation to solve for n, the number of moles. This gives us n = PV/RT. Plugging in the given values, we get n = (1.0 atm * 2.0 L + 1.0 atm * 1.0 L) / (0.0821 L atm/mol K * 298 K) = 0.0482 mol A + 0.0241 mol B. Next, we need to determine the limiting reactant, the reactant that will be completely used up in the reaction. To do this, we compare the number of moles of each gas present to the coefficients in the balanced chemical equation. It can be seen that for every 2 moles of A, 1 mole of B is required for the reaction to take place. Since there are more moles of A initially, we can conclude that B is the limiting reactant. Using the stoichiometric ratio, we can determine the number of moles of C produced. For every 1 mole of B, 2 moles of C will be produced. This means that after the reaction, the final number of moles of C will be 0.0241 mol B * 2 mol C/1 mol B = 0.0482 mol C. To find the final pressure of the flask, we use the same equation as before, PV = nRT, and solve for P. Plugging in the values, we get P = (0.0482 mol C * 0.0821 L atm/mol K * 298 K) / 3 L = 0.659 atm. Therefore, the pressure after the reaction goes to completion will be approximately 0.67 atm.