For each of the following unbalanced equations, suppose that exactly 15.0g of each reactant is taken. Determine which reactant is limiting, and calculate what mass of each product is expected. assume the limiting reactant is completely consumed.
a)Al+HCL->AlCl3+H2
b)NaOH+CO2->Na2CO3+H2O
c)Pb(NO3)2+HCL->PbCl2+HNO3
d)K+I2->KI
Write the balanced equation.
2Al + 6HCl ==> 2AlCl3 + 3H2
Convert 15.0 g Al to moles.
Convert 15.0 g HCl to moles.
15.0 g Al x (1 mole Al/26 g) = 0.555
15.0 g HCl x (1 mole HCl/36.5 g) = 0.410
Convert moles of each to mols of either product (but stay consistent). Let's choose AlCl3.
0.555 moles Al x (2 moles AlCl3/2 moles Al) = 0.555 x 1/1 = 0.555 moles AlCl3.
0.410 moles HCl x (2 moles AlCl3/6 moles HCl) = 0.410 x 2/6 = 0.137 moles AlCl3.
The SMALLER number of moles produced by the HCl means two things: (a) HCl is the limiting reagent, and (b)0.137 moles AlCl3 is the amount of product formed using 15.0 g of each reactant initially.
Now convert moles AlCl3 to grams.
0.137 moles AlCl3 x (133.5 g AlCl3/1 mole AlCl3) = 18.3 g.
All of the problems are worked this way. Just follow this template. And you need to redo this one from the beginning for two reasons. First, I estimated the atomic masses and molar masses and you need to use the exact values to obtain a correct answer. Second, you need the practice and this can be a good practice problem because you know the answers for each step.
To determine the limiting reactant and calculate the mass of each product, we need to convert the given mass of reactant to the number of moles and then compare the moles of each reactant.
a) Al + HCl -> AlCl3 + H2
First, we calculate the number of moles of Al and HCl using their molar masses:
Number of moles of Al = 15.0g / molar mass of Al
Number of moles of HCl = 15.0g / molar mass of HCl
Next, we use the stoichiometric coefficients from the balanced equation to compare the moles of Al and HCl. The balanced equation tells us that the stoichiometric ratio of Al to HCl is 1:3.
If the moles of Al are less than the moles of HCl, then Al is the limiting reactant. Otherwise, HCl is the limiting reactant.
Once we determine the limiting reactant, we can calculate the mass of each product using the stoichiometry of the balanced equation.
b) NaOH + CO2 -> Na2CO3 + H2O
c) Pb(NO3)2 + HCl -> PbCl2 + HNO3
d) K + I2 -> KI
To determine the limiting reactant and calculate the mass of each product, we need to follow these steps:
Step 1: Write the balanced equation for the given unbalanced equation.
Step 2: Calculate the number of moles for each reactant using the given mass.
Step 3: Determine the stoichiometric ratio between the reactants and the products from the balanced equation.
Step 4: Identify the limiting reactant by comparing the calculated moles of each reactant.
Step 5: Calculate the theoretical yield of each product using the stoichiometric ratio and limiting reactant.
Let's go through each example using these steps:
a) Al + HCl -> AlCl3 + H2
Step 1: The balanced equation for this reaction is:
2 Al + 6 HCl -> 2 AlCl3 + 3 H2
Step 2: Calculate the moles of Al and HCl:
Molar mass of Al = 26.98 g/mol
Molar mass of HCl = 36.46 g/mol
moles of Al = mass / molar mass = 15.0 g / 26.98 g/mol = 0.556 mol
moles of HCl = mass / molar mass = 15.0 g / 36.46 g/mol = 0.411 mol
Step 3: From the balanced equation, the stoichiometric ratio of Al to HCl is 2:6 (or 1:3).
Step 4: The moles of Al is 0.556 mol, and the moles of HCl is 0.411 mol. Since the ratio is 1:3, Al is the limiting reactant.
Step 5: Using the stoichiometric ratio, the theoretical yield of AlCl3 will be:
0.556 mol Al x (2 mol AlCl3 / 2 mol Al) x (133.34 g/mol AlCl3) = 73.4 g AlCl3
The theoretical yield of H2 will be:
0.556 mol Al x (3 mol H2 / 2 mol Al) x (2.02 g/mol H2) = 1.68 g H2
b) NaOH + CO2 -> Na2CO3 + H2O
Step 1: The balanced equation for this reaction is:
2 NaOH + CO2 -> Na2CO3 + H2O
Step 2: Calculate the moles of NaOH and CO2:
Molar mass of NaOH = 39.997 g/mol
Molar mass of CO2 = 44.01 g/mol
moles of NaOH = mass / molar mass = 15.0 g / 39.997 g/mol = 0.375 mol
moles of CO2 = mass / molar mass = 15.0 g / 44.01 g/mol = 0.341 mol
Step 3: From the balanced equation, the stoichiometric ratio of NaOH to CO2 is 2:1.
Step 4: The moles of NaOH is 0.375 mol, and the moles of CO2 is 0.341 mol. Since the ratio is 2:1, CO2 is the limiting reactant.
Step 5: Using the stoichiometric ratio, the theoretical yield of Na2CO3 will be:
0.341 mol CO2 x (1 mol Na2CO3 / 1 mol CO2) x (105.99 g/mol Na2CO3) = 36.5 g Na2CO3
The theoretical yield of H2O will be:
0.341 mol CO2 x (1 mol H2O / 1 mol CO2) x (18.015 g/mol H2O) = 6.14 g H2O
c) Pb(NO3)2 + HCl -> PbCl2 + HNO3
Step 1: The balanced equation for this reaction is:
Pb(NO3)2 + 2 HCl -> PbCl2 + 2 HNO3
Step 2: Calculate the moles of Pb(NO3)2 and HCl:
Molar mass of Pb(NO3)2 = 331.21 g/mol
Molar mass of HCl = 36.46 g/mol
moles of Pb(NO3)2 = mass / molar mass = 15.0 g / 331.21 g/mol = 0.045 mol
moles of HCl = mass / molar mass = 15.0 g / 36.46 g/mol = 0.411 mol
Step 3: From the balanced equation, the stoichiometric ratio of Pb(NO3)2 to HCl is 1:2.
Step 4: The moles of Pb(NO3)2 is 0.045 mol, and the moles of HCl is 0.411 mol. Since the ratio is 1:2, Pb(NO3)2 is the limiting reactant.
Step 5: Using the stoichiometric ratio, the theoretical yield of PbCl2 will be:
0.045 mol Pb(NO3)2 x (1 mol PbCl2 / 1 mol Pb(NO3)2) x (278.11 g/mol PbCl2) = 12.5 g PbCl2
The theoretical yield of HNO3 will be:
0.045 mol Pb(NO3)2 x (2 mol HNO3 / 1 mol Pb(NO3)2) x (63.01 g/mol HNO3) = 5.67 g HNO3
d) K + I2 -> KI
Step 1: The balanced equation for this reaction is:
2 K + I2 -> 2 KI
Step 2: Calculate the moles of K and I2:
Molar mass of K = 39.098 g/mol
Molar mass of I2 = 253.80 g/mol
moles of K = mass / molar mass = 15.0 g / 39.098 g/mol = 0.383 mol
moles of I2 = mass / molar mass = 15.0 g / 253.80 g/mol = 0.059 mol
Step 3: From the balanced equation, the stoichiometric ratio of K to I2 is 2:1.
Step 4: The moles of K is 0.383 mol, and the moles of I2 is 0.059 mol. Since the ratio is 2:1, I2 is the limiting reactant.
Step 5: Using the stoichiometric ratio, the theoretical yield of KI will be:
0.059 mol I2 x (2 mol KI / 1 mol I2) x (166.00 g/mol KI) = 11.0 g KI
No additional product is expected as K is in excess.
Remember, these calculations represent the theoretical yield assuming complete and ideal conditions. In practice, the actual yield may differ due to various factors.