8. Magnesium reacts with hydrochloric acid as shown in the equation:

[tex]\[ \text{Mg} + 2\text{HCl} \longrightarrow \text{MgCl}_2 + \text{H}_2 \][/tex]

A 0.24 g sample of magnesium is added to 5.0 cm[tex]\(^3\)[/tex] of 2.0 mol/dm[tex]\(^3\)[/tex] hydrochloric acid.

(i) Which reactant, magnesium or hydrochloric acid, is in excess? Use calculations to explain your answer.

(ii) Calculate the maximum mass of magnesium chloride that can be formed in this reaction.



Answer :

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.