1) Aluminum metal reacts with chlorine gas to form solid aluminum trichloride, AlCl3. What mass of chlorine gas (Cl2) is needed to react completely with 163 g of aluminum?
2) Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium chloride and iron.
3Mg(s) + 2FeCl3(s) → 3MgCl2(s) + 2Fe(s)
A mixture of 41.0 g of magnesium ( Picture = 24.31 g/mol) and 175 g of iron(III) chloride ( Picture = 162.2 g/mol) is allowed to react. Identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete.
1.
mols Al = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols Al to mols Cl2.
Now convert mols Cl2 to g Cl2. g = mols x molar mass.
2.
Do what I did above several times; ie. as follows:
mols Mg = ?
mols Fe = ?
Using the coefficients in the balanced equation, convert mols Mg to mols Fe.
Do the same for mols FeCl3 to mols Fe.
It is likely the two values will not be the same. The correct answer in limiting reagent problems is ALWAYS the smaller number and the reagent producing that value is the limiting reagent (LR).
To determine how much of the LR was used use the coefficients as above to convert mols Mg to mols FeCl3. Then subtract amount used from the initial amount to find the amount remaining after reaction is complete.
1) To determine the mass of chlorine gas (Cl2) required to react completely with 163 g of aluminum (Al), we need to use the balanced chemical equation for the reaction:
2 Al + 3 Cl2 -> 2 AlCl3
The molar mass of aluminum is 26.98 g/mol, and the molar mass of chlorine gas is 70.90 g/mol (Cl2 has a molar mass of 35.45 g/mol since chlorine is a diatomic molecule).
First, we need to calculate the number of moles of aluminum:
Moles of aluminum = Mass of aluminum / Molar mass of aluminum
Moles of aluminum = 163 g / 26.98 g/mol
Moles of aluminum = 6.04 mol
According to the balanced equation, the molar ratio between aluminum and chlorine gas is 2:3. This means that for every 2 moles of aluminum, we need 3 moles of chlorine gas.
Using this molar ratio, we can calculate the number of moles of chlorine gas:
Moles of chlorine gas = Moles of aluminum * (3 moles of chlorine gas / 2 moles of aluminum)
Moles of chlorine gas = 6.04 mol * (3/2)
Moles of chlorine gas = 9.06 mol
Finally, we can calculate the mass of chlorine gas:
Mass of chlorine gas = Moles of chlorine gas * Molar mass of chlorine gas
Mass of chlorine gas = 9.06 mol * 70.90 g/mol
Mass of chlorine gas = 643.98 g
Therefore, the mass of chlorine gas needed to react completely with 163 g of aluminum is approximately 644 g.
2) To determine the limiting reactant and the mass of the excess reactant, we need to compare the number of moles of each reactant to their respective stoichiometric coefficients in the balanced chemical equation.
The balanced chemical equation is:
3 Mg(s) + 2 FeCl3(s) → 3 MgCl2(s) + 2 Fe(s)
First, let's calculate the number of moles for each reactant:
Moles of magnesium = Mass of magnesium / Molar mass of magnesium
Moles of magnesium = 41.0 g / 24.31 g/mol
Moles of magnesium = 1.687 mol
Moles of iron(III) chloride = Mass of iron(III) chloride / Molar mass of iron(III) chloride
Moles of iron(III) chloride = 175 g / 162.2 g/mol
Moles of iron(III) chloride = 1.079 mol
Next, we compare the moles of each reactant to their respective stoichiometric coefficients:
For magnesium, the stoichiometric coefficient is 3, and we have 1.687 moles.
For iron(III) chloride, the stoichiometric coefficient is 2, and we have 1.079 moles.
To determine the limiting reactant, we compare the moles of each reactant after dividing them by their respective stoichiometric coefficients:
Moles of magnesium per stoichiometric coefficient = Moles of magnesium / 3
Moles of magnesium per stoichiometric coefficient = 1.687 mol / 3
Moles of magnesium per stoichiometric coefficient = 0.562 mol
Moles of iron(III) chloride per stoichiometric coefficient = Moles of iron(III) chloride / 2
Moles of iron(III) chloride per stoichiometric coefficient = 1.079 mol / 2
Moles of iron(III) chloride per stoichiometric coefficient = 0.539 mol
We can see that the mole ratio of magnesium to its stoichiometric coefficient is higher than that of iron(III) chloride to its stoichiometric coefficient. Therefore, magnesium is the limiting reactant.
To calculate the mass of the excess reactant (iron(III) chloride) when the reaction is complete, we can use the mole ratio between magnesium and iron(III) chloride:
Moles of iron(III) chloride reacted with magnesium = Moles of magnesium per stoichiometric coefficient * 2
Moles of iron(III) chloride reacted with magnesium = 0.562 mol * 2
Moles of iron(III) chloride reacted with magnesium = 1.124 mol
Moles of iron(III) chloride remaining = Moles of iron(III) chloride initially - Moles of iron(III) chloride reacted with magnesium
Moles of iron(III) chloride remaining = 1.079 mol - 1.124 mol
Moles of iron(III) chloride remaining = -0.045 mol (negative value implies it is all reacted)
Therefore, there is no excess iron(III) chloride remaining when the reaction is complete.
1) To determine the mass of chlorine gas (Cl2) needed to react completely with 163 g of aluminum (Al), we can use the balanced chemical equation for the reaction:
2Al + 3Cl2 → 2AlCl3
From the equation, we can see that 2 moles of aluminum (2Al) react with 3 moles of chlorine gas (3Cl2) to produce 2 moles of aluminum trichloride (2AlCl3).
First, we need to calculate the molar mass of aluminum (Al) and chlorine gas (Cl2). The molar mass of aluminum is 26.98 g/mol and the molar mass of chlorine is 35.45 g/mol.
Next, we use the given mass of aluminum (163 g) and convert it to moles using its molar mass:
163 g Al * (1 mol Al / 26.98 g Al) = 6.040 mol Al
Since the ratio between aluminum and chlorine gas in the balanced equation is 2:3, we can calculate the moles of chlorine gas needed:
6.040 mol Al * (3 mol Cl2 / 2 mol Al) = 9.060 mol Cl2
Finally, we convert the moles of chlorine gas to mass using its molar mass:
9.060 mol Cl2 * (35.45 g Cl2 / 1 mol Cl2) = 321.7 g Cl2
Therefore, 321.7 g of chlorine gas (Cl2) is needed to react completely with 163 g of aluminum (Al).
2) To determine the limiting reactant and the mass of the excess reactant, we need to compare the amounts of magnesium (Mg) and iron(III) chloride (FeCl3) present in the mixture to the stoichiometry of the balanced equation.
The balanced equation for the reaction is:
3Mg(s) + 2FeCl3(s) → 3MgCl2(s) + 2Fe(s)
First, let's calculate the number of moles for each reactant.
For magnesium (Mg):
Mass of magnesium = 41.0 g
Molar mass of magnesium = 24.31 g/mol
Number of moles of magnesium:
41.0 g Mg * (1 mol Mg / 24.31 g Mg) = 1.686 mol Mg
For iron(III) chloride (FeCl3):
Mass of iron(III) chloride = 175 g
Molar mass of iron(III) chloride = 162.2 g/mol
Number of moles of iron(III) chloride:
175 g FeCl3 * (1 mol FeCl3 / 162.2 g FeCl3) = 1.079 mol FeCl3
Next, we can determine the stoichiometric ratio between magnesium and iron(III) chloride in the balanced equation. According to the equation, 3 moles of magnesium react with 2 moles of iron(III) chloride.
From the stoichiometry, we can calculate the number of moles of magnesium that would be required to react with the given amount of iron(III) chloride:
1.079 mol FeCl3 * (3 mol Mg / 2 mol FeCl3) = 1.619 mol Mg
Comparing this calculated amount with the actual amount of magnesium present (1.686 mol), we can see that the magnesium is in excess.
On the other hand, the iron(III) chloride is the limiting reactant since it is completely consumed in the reaction.
To calculate the mass of the excess reactant, we need to subtract the amount of the limiting reactant from the actual amount of the excess reactant.
Excess magnesium:
1.686 mol Mg - 1.619 mol Mg = 0.067 mol Mg
Mass of the excess magnesium:
0.067 mol Mg * (24.31 g Mg / 1 mol Mg) = 1.629 g Mg
Therefore, the limiting reactant is iron(III) chloride (FeCl3), and the mass of the excess reactant (magnesium) remaining in the vessel when the reaction is complete is 1.629 g.