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 $H_{2}S$, we need to use the balanced chemical equation provided:<br /><br />$Al_{2}S_{3}(s)+6H_{2}O(l)\rightarrow 2Al(OH)_{3}(s)+3H_{2}S(g)$<br /><br />From the balanced equation, we can see that 1 mole of $Al_{2}S_{3}$ reacts with 6 moles of $H_{2}O$ to produce 3 moles of $H_{2}S$. <br /><br />Given that we have 32 moles of $Al_{2}S_{3}$ and 32 moles of $H_{2}O$, we need to determine which reactant is the limiting reactant.<br /><br />Since the stoichiometric ratio between $Al_{2}S_{3}$ and $H_{2}O$ is 1:6, we can see that we have enough $H_{2}O$ to react with all the $Al_{2}S_{3}$. Therefore, $Al_{2}S_{3}$ is the limiting reactant.<br /><br />Now, we can calculate the theoretical yield of $H_{2}S$ using the limiting reactant:<br /><br />Theoretical yield of $H_{2}S = \frac{3 \text{ moles } H_{2}S}{1 \text{ mole } Al_{2}S_{3}} \times 32 \text{ moles } Al_{2}S_{3} = 96 \text{ moles } H_{2}S$<br /><br />Therefore, the correct answer is B) 96 mol $H_{2}S$.<br /><br />21. To find the mass of $NaCl$ required to make a 3.000 M solution, we need to use the formula:<br /><br />Molarity (M) = Moles of solute / Volume of solution (in liters)<br /><br />Given that the volume of the solution is 250.00 mL, we need to convert it to liters:<br /><br />250.00 mL = 0.250 L<br /><br />We also know that the molarity is 3.000 M. Rearranging the formula, we can find the moles of solute:<br /><br />Moles of solute = Molarity × Volume of solution (in liters)<br /><br />Moles of solute = 3.000 M × 0.250 L = 0.750 mol<br /><br />Now, we need to find the mass of $NaCl$ required. We can use the molar mass of $NaCl$ to calculate this:<br /><br />Molar mass of $NaCl = 22.99 \text{ g/mol } Na + 35.45 \text{ g/mol } Cl = 58.44 \text{ g/mol}$<br /><br />Mass of $NaCl = \text{Moles of solute} \times \text{Molar mass of } NaCl$<br /><br />Mass of $NaCl = 0.750 \text{ mol} \times 58.44 \text{ g/mol} = 43.83 \text{ g}$<br /><br />Therefore, the correct answer is B) 43.83 g.<br /><br />22. To find the concentration of the solution prepared by diluting 20.0 mL of a 0.200 M $LiCl$ solution to 250.0 mL, we can use the dilution formula:<br /><br />$C_1V_1 = C_2V_2$<br /><br />where $C_1$ is the initial concentration, $V_1$ is the initial volume, $C_2$ is the final concentration, and $V_2$ is the final volume.<br /><br />Given that $C_1 = 0.200 \text{ M}$, $V_1 = 20.0 \text{ mL}$, and $V_2 = 250.0 \text{ mL}$, we can solve for $C_2$:<br /><br />$C_2 = \frac{C_1V_1}{V_2} = \frac{0.200 \text{ M} \times 20.0 \text{ mL}}{250.0 \text{ mL}} = 0.0160 \text{ M}$<br /><br />Therefore, the correct answer is A) 0.0160 M.<br /><br />23. To determine the oxidation state of H in $H_{2}(g)$, we need to look at the balanced reaction provided:<br /><br />$Mg(s)+2HBr(aq)\rightarrow MgBr_{2}(aq)+H_{2}(g)$<br /><br />From the balanced equation, we can see that the oxidation state of H in $H_{2}(g)$ is 0.<br /><br />Therefore, the correct answer is C) 0.<br /><br />24. To determine the oxidizing agent in the given reaction, we need to look at the oxidation states of the elements involved:<br /><br />$2Al^{3}+(aq)+2Fe(s)\