Solid magnesium metal reacts with chlorine gas according to the following equation:

Mg + Cl2 --> MgCl2
What mass of magnesium chloride is formed in the reaction between 7 g of Mg and 7.9 g of Cl2
use the masses to two decimal places on the chart in the front of your text book. Round your answer to two decimal places, do not include units.

This is a limiting reagent problem. We know that because amounts are given for BOTH reactants.

mols Mg = grams/molar mass
mols Cl2 = grams/molar mass

Use the coefficients in the balanced equation to convert mols Mg to mols MgCl2.
Do the same to convert mols Cl2 to mols MgCl2.
It is likely that the two values will not agree which means mone of them is wrong; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent (LR).
Using the smaller value, convert to grams. g = mols x molar mass

To determine the mass of magnesium chloride formed, we need to follow these steps:

Step 1: Convert the mass of Mg and Cl2 to moles using their molar masses.
The molar mass of Mg is 24.31 g/mol. Therefore, the number of moles of Mg can be calculated as:
Moles of Mg = mass of Mg / molar mass of Mg
Moles of Mg = 7 g / 24.31 g/mol
Moles of Mg ≈ 0.29 mol (rounded to two decimal places)

The molar mass of Cl2 is 70.90 g/mol. Hence, the number of moles of Cl2 is:
Moles of Cl2 = mass of Cl2 / molar mass of Cl2
Moles of Cl2 = 7.9 g / 70.90 g/mol
Moles of Cl2 ≈ 0.11 mol (rounded to two decimal places)

Step 2: Determine the limiting reactant.
To find the limiting reactant, we compare the number of moles of each reactant by using the stoichiometry of the balanced equation. According to the chemical equation, the ratio between Mg and Cl2 is 1:1. Therefore, the reactant with the smaller number of moles is the limiting reactant.

In this case, both reactants have a 1:1 ratio of moles, so the limiting reactant is Cl2 as it has the lowest number of moles (0.11 mol).

Step 3: Calculate the moles of MgCl2 formed.
Since the reaction is 1:1 between Mg and Cl2, the moles of MgCl2 formed will be equal to the number of moles of the limiting reactant (Cl2).

Therefore, Moles of MgCl2 ≈ 0.11 mol (limiting reactant)

Step 4: Convert moles of MgCl2 to mass using its molar mass.
The molar mass of MgCl2 is 95.21 g/mol. Hence, the mass of MgCl2 formed is:
Mass of MgCl2 = Moles of MgCl2 × molar mass of MgCl2
Mass of MgCl2 = 0.11 mol × 95.21 g/mol
Mass of MgCl2 ≈ 10.47 g (rounded to two decimal places)

Therefore, the mass of magnesium chloride formed in the reaction between 7 g of Mg and 7.9 g of Cl2 is approximately 10.47 g.

To find the mass of magnesium chloride formed in the reaction, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed and thus limits the amount of product formed.

To find the limiting reactant, we can use the concept of stoichiometry. We will compare the ratio of the moles of magnesium to the moles of chlorine gas using the molar masses of the elements.

1. Determine the molar mass of magnesium (Mg):
In the periodic table, the atomic mass of magnesium (Mg) is 24.31 g/mol.

2. Calculate the number of moles of magnesium (Mg):
Moles = mass / molar mass
Moles of Mg = 7 g / 24.31 g/mol

3. Determine the molar mass of chlorine gas (Cl2):
Chlorine (Cl) atomic mass on the periodic table is 35.45 g/mol. Since there are two chlorine atoms in one molecule of chlorine gas (Cl2), the molar mass of Cl2 is 35.45 g/mol x 2 = 70.90 g/mol.

4. Calculate the number of moles of chlorine gas (Cl2):
Moles of Cl2 = 7.9 g / 70.90 g/mol

5. Compare the moles of magnesium (Mg) and chlorine gas (Cl2) using the stoichiometric ratio from the balanced equation:
The balanced equation is:
Mg + Cl2 --> MgCl2

From the equation, we see that 1 mole of Mg reacts with 1 mole of Cl2 to form 1 mole of MgCl2.

Now, compare the moles of Mg and Cl2. Whichever is lower will be the limiting reactant.

6. Determine the limiting reactant:
Compare the moles of Mg and Cl2:

Moles of Mg = (mass of Mg) / (molar mass of Mg) = 7 g / 24.31 g/mol
Moles of Cl2 = (mass of Cl2) / (molar mass of Cl2) = 7.9 g / 70.90 g/mol

Compare the moles of Mg and Cl2 to see which is lower.

7. Calculate the moles of MgCl2 formed:
Since the stoichiometric ratio is 1:1 between Mg and MgCl2, the moles of MgCl2 formed will be equal to the moles of the limiting reactant.

8. Calculate the mass of MgCl2 formed:
Mass of MgCl2 = Moles of MgCl2 x Molar mass of MgCl2

To find the molar mass of MgCl2, sum the atomic masses of magnesium (Mg) and chlorine (Cl). The atomic mass of Cl from the periodic table is 35.45 g/mol.

Molar mass of MgCl2 = (Molar mass of Mg) + 2 x (Molar mass of Cl)
Molar mass of MgCl2 = 24.31 g/mol + 2 x 35.45 g/mol

Now, calculate the mass of MgCl2 formed using the moles of the limiting reactant and the molar mass of MgCl2.

Mass of MgCl2 = Moles of MgCl2 formed x Molar mass of MgCl2

Round the final mass of MgCl2 to two decimal places, as instructed.