To determine the volume of hydrogen gas produced by the reaction of 0.020 moles of magnesium with an excess amount of hydrochloric acid at standard temperature and pressure (STP), let's follow the given chemical reaction:
[tex]\[
\text{Mg} (s) + 2 \text{HCl} (aq) \longrightarrow \text{MgCl}_2 (aq) + \text{H}_2 (g)
\][/tex]
Here is the step-by-step solution process:
1. Stoichiometry of the Reaction:
The balanced chemical equation indicates that 1 mole of magnesium (Mg) produces 1 mole of hydrogen gas (H[tex]\(_2\)[/tex]).
2. Moles of Hydrogen Gas Produced:
Since 0.020 moles of magnesium are reacting, and the ratio of Mg to H[tex]\(_2\)[/tex] is 1:1, the amount of H[tex]\(_2\)[/tex] produced will also be 0.020 moles.
3. Volume at STP:
Under standard temperature and pressure (STP) conditions, 1 mole of any ideal gas occupies 22.4 L (liters).
4. Calculate Volume of Hydrogen Gas:
[tex]\[
\text{Volume of Hydrogen Gas at STP} = \text{Moles of } H_2 \times \text{Volume of 1 Mole at STP}
\][/tex]
[tex]\[
\text{Volume of Hydrogen Gas at STP} = 0.020 \text{ moles} \times 22.4 \text{ L/mole}
\][/tex]
[tex]\[
\text{Volume of Hydrogen Gas at STP} = 0.448 \text{ L}
\][/tex]
5. Convert Liters to Milliliters:
There are 1000 milliliters (mL) in 1 liter (L).
[tex]\[
\text{Volume in milliliters} = 0.448 \text{ L} \times 1000 \text{ mL/L}
\][/tex]
[tex]\[
\text{Volume in milliliters} = 448 \text{ mL}
\][/tex]
Based on these steps, the volume of hydrogen gas produced by the reaction is:
[tex]\[
448 \, \text{mL}
\][/tex]
Therefore, the correct answer is [tex]\(448 \, \text{mL}\)[/tex].
