We are given the reaction:
[tex]\[ \text{Mg} + \text{Cl}_2 \rightarrow \text{MgCl}_2 \][/tex]
The given masses are:
- Mass of Magnesium ([tex]\( \text{Mg} \)[/tex]): 20.0 grams
- Mass of Chlorine gas ([tex]\( \text{Cl}_2 \)[/tex]): 50.0 grams
To determine the amount of magnesium that remains after the reaction, let's go through the following steps:
1. Find the molar masses of the reactants:
- Molar mass of [tex]\( \text{Mg} \)[/tex]: 24.305 g/mol
- Molar mass of [tex]\( \text{Cl}_2 \)[/tex]: 70.906 g/mol (since chlorine gas is diatomic, [tex]\(2 \times 35.453 \text{ g/mol}\)[/tex])
2. Calculate the moles of each reactant:
- Moles of [tex]\( \text{Mg} \)[/tex]:
[tex]\[
\text{Moles of } \text{Mg} = \frac{\text{Mass of Mg}}{\text{Molar mass of Mg}} = \frac{20.0 \text{ grams}}{24.305 \text{ g/mol}} \approx 0.8229 \text{ moles}
\][/tex]
- Moles of [tex]\( \text{Cl}_2 \)[/tex]:
[tex]\[
\text{Moles of } \text{Cl}_2 = \frac{\text{Mass of Cl}_2}{\text{Molar mass of Cl}_2} = \frac{50.0 \text{ grams}}{70.906 \text{ g/mol}} \approx 0.7052 \text{ moles}
\][/tex]
3. Determine the limiting reactant:
The stoichiometric ratio of the reaction is 1:1 (1 mol of Mg reacts with 1 mol of Cl_2). Since [tex]\( \text{Cl}_2 \)[/tex] has fewer moles ([tex]\(0.7052 \text{ moles}\)[/tex]) compared to [tex]\( \text{Mg} \)[/tex] ([tex]\(0.8229 \text{ moles}\)[/tex]), [tex]\( \text{Cl}_2 \)[/tex] is the limiting reactant.
4. Calculate the moles of [tex]\( \text{Mg} \)[/tex] that will react:
The moles of [tex]\( \text{Mg} \)[/tex] that will react with [tex]\( \text{Cl}_2 \)[/tex] is equal to the moles of [tex]\( \text{Cl}_2 \)[/tex] available, which is [tex]\(0.7052 \text{ moles}\)[/tex].
5. Calculate the mass of [tex]\( \text{Mg} \)[/tex] that reacts:
[tex]\[
\text{Mass of } \text{Mg} \text{ reacted} = \text{Moles of Mg reacted} \times \text{Molar mass of Mg} = 0.7052 \text{ moles} \times 24.305 \text{ g/mol} \approx 17.1389 \text{ grams}
\][/tex]
6. Determine the mass of [tex]\( \text{Mg} \)[/tex] that remains:
[tex]\[
\text{Mass of } \text{Mg} \text{ remaining} = \text{Initial mass of Mg} - \text{Mass of Mg reacted} = 20.0 \text{ grams} - 17.1389 \text{ grams} \approx 2.8611 \text{ grams}
\][/tex]
Therefore, at the completion of the reaction, the mass of magnesium that remains is approximately [tex]\( \boxed{2.8611 \text{ grams}} \)[/tex].
