Consider the reaction: Mg + I2 -> MgI2.
Identify the limiting reagent in of the reaction mixtures.
1. 100 atoms of Mg and 100 molecules of I2.
2. 150 atoms of Mg and 100 molecules of I2.
4. 0.16 mol Mg and 0.25 mol I2.
5. 0.14 mol Mg and 0.14 mol I2.
6. 6.078 g Mg and 63.46 g I2.
7. 1.00 g Mg and 2.00 g I2.
See the H3PO4 and CaSO4 problem above.
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To identify the limiting reagent in each reaction mixture, we need to compare the stoichiometric ratio of the reactants to determine which one will be completely consumed first, thereby limiting the amount of product that can be formed.
1. First, we need to determine how many moles of Mg and I2 are present.
100 atoms of Mg is equivalent to (100/6.022 × 10^23) mol = 1.66 × 10^-22 mol of Mg.
100 molecules of I2 is equivalent to 100 × Avogadro's number = 100 × 6.022 × 10^23 = 6.022 × 10^25 mol.
To compare the moles of Mg and I2, we divide the number of moles of each reagent by its stoichiometric coefficient:
Mg: 1.66 × 10^-22 mol / 1 mol = 1.66 × 10^-22 mol
I2: 6.022 × 10^25 mol / 1 mol = 6.022 × 10^25 mol
Since the ratio of Mg moles to I2 moles is significantly low, the limiting reagent is Mg.
2. Similar calculations can be performed to determine the moles of Mg and I2 in this reaction mixture.
Mg: 150 atoms of Mg = (150/6.022 × 10^23) mol = 2.49 × 10^-22 mol
I2: 100 molecules of I2 = 100 × 6.022 × 10^23 = 6.022 × 10^25 mol
Again, the ratio of Mg moles to I2 moles is significantly low, indicating that Mg is the limiting reagent.
3. For this option, we need to compare the number of moles of Mg and I2 given.
Mg: 0.16 mol
I2: 0.25 mol
In this case, we can directly compare the moles, and we can see that the ratio of Mg moles to I2 moles is less than 1. Therefore, Mg is the limiting reagent.
4. Here, we also compare the moles given for Mg and I2.
Mg: 0.14 mol
I2: 0.14 mol
The ratio of Mg moles to I2 moles is 1:1, meaning that they are present in stoichiometric quantities. None of the reactants are in excess or limiting.
5. Next, we convert the given masses of Mg and I2 to moles.
Mg: 6.078 g / 24.31 g/mol = 0.250 mol
I2: 63.46 g / 253.80 g/mol = 0.250 mol
Again, the ratio of Mg moles to I2 moles is 1:1, indicating that neither reactant is in excess or limiting.
6. Finally, we convert the given masses of Mg and I2 to moles.
Mg: 1.00 g / 24.31 g/mol = 0.0411 mol
I2: 2.00 g / 253.80 g/mol = 0.00789 mol
In this case, the ratio of Mg moles to I2 moles is significantly higher, indicating that I2 is the limiting reagent.
Therefore, the limiting reagents for the given reaction mixtures are:
1. Mg
2. Mg
4. Mg
5. None (both are in stoichiometric quantities)
6. None (both are in stoichiometric quantities)
7. I2
To identify the limiting reagent in a chemical reaction, we need to compare the number of moles of each reactant with the stoichiometric ratio given by the balanced equation.
The balanced equation for the reaction is:
Mg + I2 -> MgI2
In this equation, the stoichiometric ratio between magnesium (Mg) and iodine (I2) is 1:1. This means that for every 1 mole of magnesium, we need 1 mole of iodine to completely react.
Now let's analyze each reaction mixture:
1. 100 atoms of Mg and 100 molecules of I2.
To determine the moles in this reaction mixture, we need to convert the given units to moles. The atomic mass of Mg is approximately 24.31 g/mol. Thus, 100 atoms of Mg would be (100 atoms) / (6.022 × 10^23 atoms/mol) ≈ (100/6.022 × 10^23) mol ≈ 1.66 × 10^(-22) mol.
Since the equation requires moles, we need to convert the given molecules of I2 to moles as well. The molar mass of I2 is approximately 253.81 g/mol. Thus, 100 molecules of I2 would be (100 molecules) / (6.022 × 10^23 molecules/mol) ≈ (100/6.022 × 10^23) mol ≈ 1.66 × 10^(-22) mol.
From the stoichiometric ratio of the balanced equation, we see that both Mg and I2 are present in equal amounts in this reaction mixture. Therefore, neither reactant is limiting, and the answer is None.
2. 150 atoms of Mg and 100 molecules of I2.
Following the same process as above, we can calculate the number of moles for each reactant. 150 atoms of Mg would be approximately 2.49 × 10^(-22) mol, whereas 100 molecules of I2 would also be approximately 1.66 × 10^(-22) mol. Again, both reactants are present in equal amounts, so the answer is None.
3. 0.16 mol Mg and 0.25 mol I2.
Based on the stoichiometric ratio of the balanced equation, 1 mole of Mg reacts with 1 mole of I2. In this case, both reactants have the proper stoichiometric ratio, so the answer is None.
4. 0.14 mol Mg and 0.14 mol I2.
Since the stoichiometric ratio is 1 mole of Mg to 1 mole of I2, the amounts of both reactants are in the correct ratio. Therefore, the answer is None.
5. 6.078 g Mg and 63.46 g I2.
To determine the number of moles, we need to divide the given masses by the molar masses of each substance. The molar mass of Mg is approximately 24.31 g/mol, so (6.078 g Mg) / (24.31 g/mol) ≈ 0.25 mol Mg. The molar mass of I2 is approximately 253.81 g/mol, so (63.46 g I2) / (253.81 g/mol) ≈ 0.25 mol I2.
Both reactants have equal amounts, so the answer is None.
6. 1.00 g Mg and 2.00 g I2.
Again, we need to convert the given masses to moles using the molar masses of each substance. (1.00 g Mg) / (24.31 g/mol) ≈ 0.041 mol Mg, and (2.00 g I2) / (253.81 g/mol) ≈ 0.009 mol I2.
Comparing the number of moles, we see that there is less I2 present (0.009 mol) than the stoichiometric ratio (0.041 mol of Mg). Therefore, iodine (I2) is the limiting reagent.
Based on the analysis, the limiting reagent is I2 in reaction mixture 7.