Sure, let's break down the problem step-by-step:
### Step 1: Calculate the moles of magnesium (Mg) used.
Given data:
- Mass of Mg = 0.24 g
- Molar mass of Mg = 24.305 g/mol
To find the moles of Mg:
[tex]\[ \text{Moles of Mg} = \frac{\text{Mass of Mg}}{\text{Molar mass of Mg}} \][/tex]
[tex]\[ \text{Moles of Mg} = \frac{0.24 \text{ g}}{24.305 \text{ g/mol}} \][/tex]
[tex]\[ \text{Moles of Mg} \approx 0.009874 \text{ moles} \][/tex]
### Step 2: Calculate the moles of hydrochloric acid (HCl) used.
Given data:
- Volume of HCl = 5.0 cm³
- Concentration of HCl = 2.0 mol/dm³
Convert the volume from cm³ to dm³:
[tex]\[ \text{Volume of HCl in dm}^3 = 5.0 \text{ cm}^3 \times \frac{1 \text{ dm}^3}{1000 \text{ cm}^3} = 0.005 \text{ dm}^3 \][/tex]
To find the moles of HCl:
[tex]\[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} \][/tex]
[tex]\[ \text{Moles of HCl} = 2.0 \text{ mol/dm}^3 \times 0.005 \text{ dm}^3 \][/tex]
[tex]\[ \text{Moles of HCl} = 0.01 \text{ moles} \][/tex]
### Step 3: Determine the stoichiometric relationship.
The reaction is:
[tex]\[ Mg + 2HCl \rightarrow MgCl_2 + H_2 \][/tex]
From the stoichiometry, 1 mole of Mg reacts with 2 moles of HCl. Thus:
[tex]\[ \text{Moles of HCl needed} = \text{Moles of Mg} \times 2 \][/tex]
[tex]\[ \text{Moles of HCl needed} = 0.009874 \text{ moles} \times 2 \approx 0.01975 \text{ moles} \][/tex]
### Step 4: Compare moles of HCl available with moles of HCl needed.
- Moles of HCl available = 0.01 moles
- Moles of HCl needed = 0.01975 moles
Since 0.01 moles of HCl is less than 0.01975 moles, HCl is the limiting reactant, and magnesium (Mg) is in excess.
Answer to (i):
Hydrochloric acid (HCl) is the limiting reactant, and magnesium (Mg) is in excess.
### Step 5: Calculate the maximum mass of magnesium chloride (MgCl₂) formed.
Given data:
- Moles of the limiting reactant (HCl) = 0.01 moles
Since HCl is limiting, we first need to find out how many moles of Mg are reacting:
[tex]\[ \text{Moles of Mg reacting} = \frac{\text{Moles of HCl}}{2} \][/tex]
[tex]\[ \text{Moles of Mg reacting} = \frac{0.01}{2} = 0.005 \text{ moles} \][/tex]
From the reaction, 1 mole of Mg forms 1 mole of MgCl₂, so:
[tex]\[ \text{Moles of MgCl}_2 = \text{Moles of Mg reacting} = 0.005 \text{ moles} \][/tex]
Given the molar mass of MgCl₂ = 95.211 g/mol, the mass of MgCl₂ formed can be calculated as:
[tex]\[ \text{Mass of MgCl}_2 = \text{Moles of MgCl}_2 \times \text{Molar mass of MgCl}_2 \][/tex]
[tex]\[ \text{Mass of MgCl}_2 = 0.005 \text{ moles} \times 95.211 \text{ g/mol} \][/tex]
[tex]\[ \text{Mass of MgCl}_2 \approx 0.476 \text{ g} \][/tex]
Answer to (ii):
The maximum mass of magnesium chloride (MgCl₂) that can be formed is approximately 0.476 g.
