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.

  $$\text{Mg(s)} + 2\text{HCl(aq)} \to \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}$$

  From the equation, 1 mole of Mg reacts with 2 moles of HCl.

• Step 2: Calculate the moles of HCl. Given: Volume of HCl ($V$) = $250 \text{ cm}^3 = 0.250 \text{ dm}^3$ (or $0.250 \text{ L}$) Concentration of HCl ($C$) = $0.5 \text{ M}$ Moles of HCl ($n_{\text{HCl}}$) = $C \times V$

  $$n_{\text{HCl}} = 0.5 \text{ mol/L} \times 0.250 \text{ L} = 0.125 \text{ mol}$$

• Step 3: Use stoichiometry to find moles of Mg. From the balanced equation, the mole ratio of Mg to HCl is $1:2$.

  $$n_{\text{Mg}} = n_{\text{HCl}} \times \frac{1 \text{ mol Mg}}{2 \text{ mol HCl}}$$ $$n_{\text{Mg}} = 0.125 \text{ mol HCl} \times \frac{1}{2} = 0.0625 \text{ mol}$$

• Step 4: Convert moles of Mg to mass of Mg. Molar mass of Mg = $24 \text{ g/mol}$ Mass of Mg = $n_{\text{Mg}} \times \text{Molar mass of Mg}$

  $$\text{Mass of Mg} = 0.0625 \text{ mol} \times 24 \text{ g/mol} = 1.5 \text{ g}$$

The mass of magnesium required is 1.5 g.
The correct option is B) 1.5g.

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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.

  $$\text{NaOH(aq)} + \text{HCl(aq)} \to \text{NaCl(aq)} + \text{H}_2\text{O(l)}$$

  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) = $39.45 \text{ cm}^3$ $C_b$ = concentration of base (NaOH) = $0.15 \text{ M}$ $V_b$ = volume of base (NaOH) = $24.83 \text{ cm}^3$

• Step 3: Solve for $C_a$.

  $$C_a = \frac{C_b V_b}{V_a}$$ $$C_a = \frac{0.15 \text{ M} \times 24.83 \text{ cm}^3}{39.45 \text{ cm}^3}$$ $$C_a = \frac{3.7245}{39.45} \text{ M}$$ $$C_a \approx 0.0944 \text{ M}$$

The molarity of the HCl is approximately 0.094 M.
The correct option is A) 0.094M.

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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 ($V$) = $0.175 \text{ dm}^3 = 0.175 \text{ L}$ Concentration of H2SO4 ($C$) = $6.00 \text{ M}$ Moles of H2SO4 ($n_{\text{H}_2\text{SO}_4}$) = $C \times V$

  $$n_{\text{H}_2\text{SO}_4} = 6.00 \text{ mol/L} \times 0.175 \text{ L} = 1.05 \text{ mol}$$

• 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 = $2.016 + 32.06 + 63.996 = 98.072 \text{ g/mol}$ (Using common rounded values: H=1, S=32, O=16 gives $2(1) + 32 + 4(16) = 2 + 32 + 64 = 98 \text{ g/mol}$) Let's use 98.07 g/mol for better precision.

• Step 3: Convert moles of H2SO4 to mass of H2SO4. Mass of H2SO4 = $n_{\text{H}_2\text{SO}_4} \times \text{Molar mass of H}_2\text{SO}_4$

  $$\text{Mass of H}_2\text{SO}_4 = 1.05 \text{ mol} \times 98.07 \text{ g/mol} = 102.9735 \text{ g}$$

  Rounding to one decimal place, this is approximately $103.0 \text{ g}$.

The mass of H2SO4 necessary is 103.0 g.
The correct option is B) 103.0g.

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4. What volume of 1.5M solution of KOH would contain 0.045 moles?

• Step 1: Identify knowns and unknowns. Concentration of KOH ($C$) = $1.5 \text{ M}$ Moles of KOH ($n$) = $0.045 \text{ moles}$ Volume of KOH ($V$) = ?

• Step 2: Use the concentration formula to find volume in liters. Moles = Concentration $\times$ Volume (in L)

  $$V \text{ (in L)} = \frac{n}{C}$$ $$V \text{ (in L)} = \frac{0.045 \text{ mol}}{1.5 \text{ mol/L}}$$ $$V \text{ (in L)} = 0.03 \text{ L}$$

• Step 3: Convert volume from L to cm³. $1 \text{ L} = 1000 \text{ cm}^3$

  $$V \text{ (in cm}^3) = 0.03 \text{ L} \times 1000 \text{ cm}^3/\text{L}$$ $$V \text{ (in cm}^3) = 30 \text{ cm}^3$$

The volume of 1.5M KOH solution that would contain 0.045 moles is 30.00 cm³.
The correct option is B) 30.00cm³.