Certainly! Let's solve this problem step-by-step:
### Step 1: Determine moles of HCl
Given:
- Volume of HCl solution = 250 mL = 0.250 L (since 1 L = 1000 mL)
- Molarity of HCl solution = 1.0 M (mol/L)
Moles of HCl ([tex]\(n_{HCl}\)[/tex]) can be calculated using the formula:
[tex]\[
n_{HCl} = Molarity \times Volume
\][/tex]
[tex]\[
n_{HCl} = 1.0 \, \text{M} \times 0.250 \, \text{L} = 0.25 \, \text{mol}
\][/tex]
### Step 2: Use stoichiometry to find moles of Mg required
According to the balanced chemical equation:
[tex]\[
2 \, \text{HCl (aq)} + \text{Mg(s)} \rightarrow \text{MgCl}_2 \, \text{(aq)} + \text{H}_2 \, \text{(g)}
\][/tex]
We see that 2 moles of HCl react with 1 mole of Mg.
This indicates that for every 2 moles of HCl, 1 mole of Mg is required. Therefore, the ratio of HCl to Mg is 2:1.
To find the moles of Mg required for 0.25 mol HCl, we use the stoichiometric ratio:
[tex]\[
\text{Moles of Mg required} = \frac{0.25 \, \text{mol HCl}}{2} = 0.125 \, \text{mol Mg}
\][/tex]
### Step 3: Calculate the mass of Mg required
Given:
- Molar mass of Mg = 24.31 g/mol
We can find the mass of Mg required using the formula:
[tex]\[
\text{Mass of Mg} = \text{Moles of Mg} \times \text{Molar mass of Mg}
\][/tex]
[tex]\[
\text{Mass of Mg} = 0.125 \, \text{mol} \times 24.31 \, \text{g/mol} = 3.03875 \, \text{g}
\][/tex]
### Conclusion
The mass of Mg required to react completely with 250 mL of 1.0 M HCl is [tex]\(3.03875 \, \text{g}\)[/tex].
Hence, 0.25 mol HCl would react with 0.125 mol Mg, and the mass of Mg required is 3.03875 grams.
