How much of magnesium is required to react with 250cm³ of 0.5 M HCl [Mg = 24]

Here are the solutions to the problems: 1. How much of magnesium is required to react with 250cm³ of 0.5 M HCl [Mg = 24] Step 1: Write the balanced chemical equation. Magnesium reacts with hydrochloric acid to produce magnesium chloride and hydrogen gas. Mg(s) + 2HCl(aq) MgCl_2(aq) + H_2(g) From the equation, 1 mole of Mg reacts with 2 moles of HCl. Step 2: Calculate the moles of HCl. Given: Volume of HCl (V) = 250 cm^3 = 0.250 dm^3 (or 0.250 L) Concentration of HCl (C) = 0.5 M Moles of HCl (n_HCl) = C × V n_HCl = 0.5 mol/L × 0.250 L = 0.125 mol Step 3: Use stoichiometry to find moles of Mg. From the balanced equation, the mole ratio of Mg to HCl is 1:2. n_Mg = n_HCl × 1 mol Mg2 mol HCl n_Mg = 0.125 mol HCl × (1)/(2) = 0.0625 mol Step 4: Convert moles of Mg to mass of Mg. Molar mass of Mg = 24 g/mol Mass of Mg = n_Mg × Molar mass of Mg Mass of Mg = 0.0625 mol × 24 g/mol = 1.5 g The mass of magnesium required is 1.5 g. The correct option is B) 1.5g. --- 2. If 24.83cm³ of 0.15M NaOH is titrated to its end point with 39.45cm³ of HCl. What is the molarity of the HCl Step 1: Write the balanced chemical equation. Sodium hydroxide reacts with hydrochloric acid in a neutralization reaction. NaOH(aq) + HCl(aq) NaCl(aq) + H_2O(l) From the equation, 1 mole of NaOH reacts with 1 mole of HCl. Step 2: Use the titration formula. For a 1:1 mole ratio reaction (n_a = n_b = 1), the titration formula is: C_a V_a = C_b V_b Where: C_a = concentration of acid (HCl) = ? V_a = volume of acid (HCl) = 39.45 cm^3 C_b = concentration of base (NaOH) = 0.15 M V_b = volume of base (NaOH) = 24.83 cm^3 Step 3: Solve for C_a. C_a = (C_b V_b)/(V_a) C_a = 0.15 M × 24.83 cm^339.45 cm^3 C_a = (3.7245)/(39.45) M C_a ≈ 0.0944 M The molarity of the HCl is approximately 0.094 M. The correct option is A) 0.094M. --- 3. How many grams of H2SO4 are necessary for the preparation of 0.175 dm³ of 6.00M H2SO4? Step 1: Calculate the moles of H2SO4. Given: Volume of H2SO4 (V) = 0.175 dm^3 = 0.175 L Concentration of H2SO4 (C) = 6.00 M Moles of H2SO4 (n_H_2SO_4) = C × V n_H_2SO_4 = 6.00 mol/L × 0.175 L = 1.05 mol Step 2: Calculate the molar mass of H2SO4. Atomic masses: H = 1.008 g/mol, S = 32.06 g/mol, O = 15.999 g/mol Molar mass of H2SO4 = 2(1.008) + 32.06 + 4(15.999) Molar mass of H2SO4 = 2.016 + 32.06 + 63.996 = 98.072 g/mol (Using common rounded values: H=1, S=32, O=16 gives 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98 g/mol) Let's use 98.07 g/mol for better precision. Step 3: Convert moles of H2SO4 to mass of H2SO4. Mass of H2SO4 = n_H_2SO_4 × Molar mass of H_2SO_4 Mass of H_2SO_4 = 1.05 mol × 98.07 g/mol = 102.9735 g Rounding to one decimal place, this is approximately 103.0 g. The mass of H2SO4 necessary is 103.0 g. The correct option is B) 103.0g. --- 4. What volume of 1.5M solution of KOH would contain 0.045 moles? Step 1: Identify knowns and unknowns. Concentration of KOH (C) = 1.5 M Moles of KOH (n) = 0.045 moles Volume of KOH (V) = ? Step 2: Use the concentration formula to find volume in liters. Moles = Concentration × Volume (in L) V (in L) = (n)/(C) V (in L) = 0.045 mol1.5 mol/L V (in L) = 0.03 L Step 3: Convert volume from L to cm³. 1 L = 1000 cm^3 V (in cm^3) = 0.03 L × 1000 cm^3/L V (in cm^3) = 30 cm^3 The volume of 1.5M KOH solution that would contain 0.045 moles is 30.00 cm³. The correct option is B) 30.00cm³.