For the reaction 2Na (s) + Cl2 (g) --> 2NaCl (s), how many grams of NaCl could be produced from 103.0 g of Na and 13.0 L of Cl2 (at STP)? What is the limiting reactant?
To find out how many grams of NaCl could be produced, we need to determine the limiting reactant first. The limiting reactant is the reactant that is fully consumed, limiting the amount of product that can be formed.
Let's calculate the moles of Na and Cl2:
Molar mass of Na = 22.99 g/mol
Molar mass of Cl2 = 35.45 g/mol
Number of moles of Na:
moles of Na = mass of Na / molar mass of Na
moles of Na = 103.0 g / 22.99 g/mol
moles of Na ≈ 4.48 mol
Number of moles of Cl2:
Using the ideal gas law, PV = nRT, we can calculate the number of moles of Cl2 at STP (Standard Temperature and Pressure):
P = 1 atm
V = 13.0 L
R = 0.0821 L·atm/(mol·K)
T = 273.15 K
n = PV / RT
n = (1 atm) (13.0 L) / (0.0821 L·atm/(mol·K) * 273.15 K)
n ≈ 0.553 mol
From the balanced equation, we can see that the mole ratio of Na to NaCl is 2:2 or 1:1. Therefore, 1 mole of Na reacts to form 1 mole of NaCl.
Since we have 4.48 moles of Na and 0.553 moles of Cl2, the limiting reactant is Cl2 because it is present in a smaller amount.
Now, let's calculate the amount of NaCl produced:
From the balanced equation, we know that 2 moles of NaCl are formed for every 1 mole of Cl2.
moles of NaCl = 2 * (moles of Cl2)
moles of NaCl = 2 * 0.553 mol
moles of NaCl ≈ 1.11 mol
Finally, let's calculate the grams of NaCl:
mass of NaCl = moles of NaCl * molar mass of NaCl
mass of NaCl = 1.11 mol * 58.44 g/mol
mass of NaCl ≈ 64.67 g
Therefore, approximately 64.67 grams of NaCl could be produced from the given amounts of Na and Cl2. The limiting reactant is Cl2.
To find out how many grams of NaCl could be produced and determine the limiting reactant, we need to follow these steps:
1. Convert the mass of Na (in grams) to moles using its molar mass.
2. Convert the volume of Cl2 (in liters) to moles using the ideal gas law at standard temperature and pressure (STP).
3. Set up the balanced equation to compare the stoichiometry of the reactants and products.
4. Determine the limiting reactant by comparing the number of moles of each reactant to the stoichiometric ratio in the balanced equation.
5. Use the limiting reactant to calculate the maximum theoretical yield of NaCl.
Now let's execute these steps:
Step 1: Convert the mass of Na to moles.
The molar mass of Na is 22.99 g/mol.
Calculating the number of moles of Na:
Number of moles of Na = Mass of Na / Molar mass of Na
Number of moles of Na = 103.0 g / 22.99 g/mol
Step 2: Convert the volume of Cl2 to moles using the ideal gas law.
At STP, 1 mole of any gas occupies 22.4 liters.
Calculating the number of moles of Cl2:
Number of moles of Cl2 = Volume of Cl2 / 22.4 L/mol
Number of moles of Cl2 = 13.0 L / 22.4 L/mol
Step 3: Write and balance the chemical equation.
The balanced equation for the reaction is:
2Na (s) + Cl2 (g) --> 2NaCl (s)
Step 4: Determine the limiting reactant.
To find the limiting reactant, we need to compare the ratio of moles of Na and Cl2 to the stoichiometric ratio in the balanced equation.
From step 1, we have the number of moles of Na.
From step 2, we have the number of moles of Cl2.
The stoichiometric ratio from the balanced equation is 2:1 for Na:Cl2.
Let's calculate the number of moles of NaCl that can be produced using both the moles of Na and Cl2.
With Na: Number of moles of NaCl = Number of moles of Na * (2 moles of NaCl / 2 moles of Na)
With Cl2: Number of moles of NaCl = Number of moles of Cl2 * (2 moles of NaCl / 1 mole of Cl2)
Step 5: Determine the maximum theoretical yield of NaCl.
The limiting reactant is the one that produces the lesser amount of NaCl.
Now, compare the calculated number of moles of NaCl from both reactants.
If the number of moles of NaCl obtained from Na is smaller, Na is the limiting reactant.
If the number of moles of NaCl obtained from Cl2 is smaller, Cl2 is the limiting reactant.
After identifying the limiting reactant, calculate the maximum theoretical yield of NaCl using the moles of the limiting reactant.
Finally, convert the moles of NaCl to grams using its molar mass.
I hope this explanation helps you understand the process of finding the limiting reactant and calculating the maximum theoretical yield of NaCl.
Convert 103.0 g Na to moles. moles = grams/molar mass
Convert 13.0L Cl2 to moles . moles = L/22.4
Using the coefficients in the balanced equation, convert moles Na to moles NaCl.
Do the same for moles Cl2 to moles NaCl.
You will obtain two different answers and both can't be right. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing the value is the limiting reagent.
grams NaCl = moles NaCl x molar mass NaCl.