in you a. 310 d. 120 b. 87 c. 200 e. none 15 One mole of two dilferent subitances A and have equel __ a. Nmber of portide d. moleolar mat b. molor mas e. pertentoge composition of the elements c. formula mas Part IV: Problem.selving questions. Solve the following problem approprintely. 16. Colculate the molenter the following compound a. Nipher dioxide (SO_(2)) a gar that is responsible for ocid roin end b. coffeine [C_(1)H_(4)O_(2)]. o Mindant present h tea, collee, and talo beveroges 17. How many moles of CO_(2) are present in 176 o of CO_(2) . 18. How many grom of exygen can be prepored by the decomposition of 12 grami of meroury (ii) caldel 19. 25 of NH_(3) is mixed wh 4 moles of O_(2) n the given readion 4NH_(3)(g)+5O_(2)(g)arrow 4NO(g)+6H_(2)O(l) a Which is the limiting reoctart b. What matiol NO is formed c. What mou of H_(2)O h formed?

Let's go through each question step by step:<br /><br />15. One mole of two different substances A and B have equal:<br />a. Number of particles<br />b. Molecular mass<br />c. Formula mass<br />d. Molar mass<br />e. Percentage composition of the elements<br /><br />The correct answer is:<br />c. Formula mass<br /><br />Explanation: One mole of any substance contains Avogadro's number (6.022 x 10^23) of particles (atoms, molecules, or ions). The formula mass (or molecular weight) of a substance is the sum of the atomic masses of the atoms in its formula. Therefore, one mole of two different substances A and B will have the same number of particles and the same formula mass.<br /><br />16. Calculate the molar mass of the following compounds:<br />a. Sulfur dioxide (SO2)<br />b. Caffeine (C8H10N4O2)<br /><br />The correct answers are:<br />a. Sulfur dioxide (SO2): 64.06 g/mol<br />b. Caffeine (C8H10N4O2): 194.19 g/mol<br /><br />Explanation:<br />a. Sulfur dioxide (SO2) has a molar mass of 64.06 g/mol, which is the sum of the atomic masses of sulfur (32.06 g/mol) and oxygen (16.00 g/mol) multiplied by 2.<br />b. Caffeine (C8H10N4O2) has a molar mass of 194.19 g/mol, which is the sum of the atomic masses of carbon (12.01 g/mol) multiplied by 8, hydrogen (1.01 g/mol) multiplied by 10, nitrogen (14.01 g/mol) multiplied by 4, and oxygen (16.00 g/mol) multiplied by 2.<br /><br />17. How many moles of CO2 are present in 176 grams of CO2?<br /><br />The correct answer is:<br />1.44 moles<br /><br />Explanation:<br />To calculate the number of moles of CO2, we need to divide the given mass of CO2 by its molar mass. The molar mass of CO2 is 44.01 g/mol. Therefore, the number of moles of CO2 is 176 g / 44.01 g/mol = 4.00 moles.<br /><br />18. How many grams of oxygen can be produced by the decomposition of 12 grams of mercury (II) oxide?<br /><br />The correct answer is:<br />8.00 grams<br /><br />Explanation:<br />The balanced chemical equation for the decomposition of mercury (II) oxide (HgO) is:<br />2HgO(s) → 2Hg(l) + O2(g)<br /><br />From the equation, we can see that 2 moles of HgO produce 1 mole of O2. The molar mass of HgO is 216.59 g/mol. Therefore, 12 g of HgO is equivalent to 12 g / 216.59 g/mol = 0.0556 moles of HgO. Since 2 moles of HgO produce 1 mole of O2, 0.0556 moles of HgO will produce 0.0278 moles of O2. The molar mass of O2 is 32.00 g/mol, so the mass of oxygen produced is 0.0278 moles * 32.00 g/mol = 0.8904 grams.<br /><br />19. 25 grams of NH3 is mixed with moles of O2 in the given reaction:<br />4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)<br /><br />a. Which is the limiting reactant?<br />b. What mass of NO is formed?<br />c. What mass of H2O is formed?<br /><br />The correct answers are:<br />a. NH3 is the limiting reactant.<br />b. 16.00 grams of NO is formed.<br />c. 18.00 grams of H2O is formed.<br /><br />Explanation:<br />a. To determine the limiting reactant, we need to compare the mole ratio of NH3 to O2 with the stoichiometric ratio in the balanced chemical equation. The molar mass of NH3 is 17.03 g/mol, so 25 g of NH3 is equivalent to 1.46 moles of NH3. The molar mass of O2 is 32.00 g/mol, so 1 mole of O2 is equivalent to 32.00 g of O2. The stoichiometric ratio of NH3 to O2 is 4:5, so the mole ratio of NH3 to O2 should be 4:5. Since the mole ratio of NH3 to O2 is less than the stoichiometric ratio, NH3 is the limiting reactant.<br />b. Using the balanced chemical equation, we can see that 4 moles of NH3 produce 4 moles of NO. Therefore, 1