2Sb + 3I2->2SbI3
Determine the limiting reactant and theoretical yield when 1.2 mol of Sb and 2.40 mol I2 are mixed.
There are shorter ways of doing this but I like the following method.
Convert 1.2 mol Sb to mols SbI3.
Convert 2.4 mol I2 to mols SbI3.
The smaller number will be the correct one and that will be the limiting reagent. The theoretical yield will be that number of mols SbI3 x molar mass SbI3.
Post your work if you need more help.
56.23
To determine the limiting reactant and theoretical yield, we need to compare the moles of each reactant to the stoichiometry of the balanced equation.
The balanced equation is: 2Sb + 3I2 -> 2SbI3
The stoichiometry of the balanced equation tells us that 2 moles of Sb react with 3 moles of I2 to form 2 moles of SbI3.
First, let's calculate the number of moles for each reactant:
For Sb: 1.2 mol
For I2: 2.40 mol
Next, we need to determine which reactant is the limiting reactant. The limiting reactant is the one that is completely consumed during the reaction, which means it will determine the maximum amount of product that can be formed.
Using the stoichiometry, we can calculate the number of moles of SbI3 that can be formed from the given moles of each reactant.
For Sb: 1.2 mol Sb * (2 mol SbI3 / 2 mol Sb) = 1.2 mol SbI3
For I2: 2.40 mol I2 * (2 mol SbI3 / 3 mol I2) = 1.60 mol SbI3
Since the stoichiometry tells us that 2 moles of Sb are required to react with 3 moles of I2, and we have an excess of I2, we can only form 1.20 mol of SbI3 from the available 1.2 mol of Sb. Therefore, Sb is the limiting reactant.
Now, let's calculate the theoretical yield of SbI3 using the limiting reactant:
For Sb: 1.2 mol Sb * (2 mol SbI3 / 2 mol Sb) = 1.2 mol SbI3 (theoretical yield)
Therefore, the limiting reactant is Sb and the theoretical yield of SbI3 is 1.2 mol.
To determine the limiting reactant and theoretical yield, you need to compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation.
The balanced equation for the reaction is:
2Sb + 3I2 -> 2SbI3
First, calculate the number of moles of each reactant:
Moles of Sb = 1.2 mol
Moles of I2 = 2.40 mol
Next, compare the moles of each reactant to the stoichiometric ratio in the balanced equation:
From the balanced equation, you can see that the ratio between Sb and I2 is 2:3. So for every 2 moles of Sb, you need 3 moles of I2.
Moles of Sb required = (2/3) * moles of I2
Moles of Sb required = (2/3) * 2.40 mol
Moles of Sb required = 1.60 mol
Now, compare the moles of Sb available (1.2 mol) to the moles of Sb required (1.60 mol):
Since the moles of Sb available (1.2 mol) is less than the moles of Sb required (1.60 mol), Sb is the limiting reactant.
To calculate the theoretical yield, you need to know the molar mass of SbI3. The molar mass of SbI3 is 419.52 g/mol.
The balanced equation indicates that 2 moles of SbI3 are produced for every 2 moles of Sb.
Moles of SbI3 produced = (2/2) * moles of Sb
Moles of SbI3 produced = (2/2) * 1.20 mol
Moles of SbI3 produced = 1.20 mol
Finally, calculate the theoretical yield in grams:
Theoretical yield = Moles of SbI3 produced * Molar mass of SbI3
Theoretical yield = 1.20 mol * 419.52 g/mol
Theoretical yield = 503.42 g
Therefore, the limiting reactant is Sb, and the theoretical yield is 503.42 grams of SbI3.