20. Determine the theoretical yield of H_(2)S (in moles) if 32 mol Al_(2)S_(3) and 32 mol H_(2)O are reacted according to the following balanced reaction. A possibly useful molar mass is Al_(2)S_(3)=150.17g/mol Al_(2)S_(3)(s)+6H_(2)O(l)arrow 2Al(OH)_(3)(s)+3H_(2)S(g) A) 16molH_(2)S B) 96 mol H_(2)S C) 48 mol H2S D) 32molH_(2)S E) 64molH_(2)S 21. How many grams of NaCl are required to make 250.00 mL of a 3.000 M solution? Na:22.99g/mol CI: 35.45g/mol) A) 58.40 g B) 43.83 g C) 175.3 g D) 14.60 g E) 128 g 22. What is the concentration of a solution prepared by diluting 20.0 ml.of a 0.200 M LiCl to 250.0mL A) 0.0160 M B) 0.160 M C) 2.50 M D) 0.00800 M E) 0.0320 M 23. What is the oxidation :tate of H in H_(2)(g) in th: following belanced reaction? Mg(s)+2HBr(aq)arrow MgBr_(2)(aq)+H_(2)(g) C)0 D)-1 E)-2 A) +1 B) +2 24. What is the oxidizing agent in the following reaction? 2Al^3+(aq)+2Fe(s)arrow 2Al(s)+3Fe^2+(aq) E) none A) Al^3+ B) Fe C) Al D) Fe^2+ 25. Determine the limiting reactant (LR) and the mass (in g)of nitrogen that can be formed from 50.0 g N_(2)O_(4) and 45.0gN_(2)H_(4) from the unbalanced reaction given below. Some possibly useful atomic and H=I molarmassolarest slowes N=14.01g/mol.N_(2)H_(4)=32.05g/mol molarmasses_(2)ast s_(2)follows N_(2)O_(4)=9 A) LR=N_(2)H_(4),59.0gN_(2) formed C) LR=N_(2)H_(4),13.3gN_(2)formed D LR=N_(2)O_(4),105gN_(2)formed E) No LR, 45.0gN_(2) formed

20. To determine the theoretical yield of , we need to use the balanced chemical equation provided: From the balanced equation, we can see that 1 mole of reacts with 6 moles of to produce 3 moles of . Given that we have 32 moles of and 32 moles of , we need to determine which reactant is the limiting reactant.Since the stoichiometric ratio between and is 1:6, we can see that we have enough to react with all the . Therefore, is the limiting reactant.Now, we can calculate the theoretical yield of using the limiting reactant:Theoretical yield of Therefore, the correct answer is B) 96 mol .21. To find the mass of required to make a 3.000 M solution, we need to use the formula:Molarity (M) = Moles of solute / Volume of solution (in liters)Given that the volume of the solution is 250.00 mL, we need to convert it to liters:250.00 mL = 0.250 LWe also know that the molarity is 3.000 M. Rearranging the formula, we can find the moles of solute:Moles of solute = Molarity × Volume of solution (in liters)Moles of solute = 3.000 M × 0.250 L = 0.750 molNow, we need to find the mass of required. We can use the molar mass of to calculate this:Molar mass of Mass of Mass of Therefore, the correct answer is B) 43.83 g.22. To find the concentration of the solution prepared by diluting 20.0 mL of a 0.200 M solution to 250.0 mL, we can use the dilution formula: where is the initial concentration, is the initial volume, is the final concentration, and is the final volume.Given that , , and , we can solve for : Therefore, the correct answer is A) 0.0160 M.23. To determine the oxidation state of H in , we need to look at the balanced reaction provided: From the balanced equation, we can see that the oxidation state of H in is 0.Therefore, the correct answer is C) 0.24. To determine the oxidizing agent in the given reaction, we need to look at the oxidation states of the elements involved:$2Al^{3}+(aq)+2Fe(s)\