[tex]\[
Mg + Cl_2 \rightarrow MgCl_2
\][/tex]

When 20.0 grams of solid magnesium react with 50.0 grams of chlorine gas, the chlorine gas is the limiting reactant. What mass of magnesium remains at the completion of the reaction?

[tex]\[
\text{[?] g Mg remain}
\][/tex]



Answer :

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].