If 40g of Sodium (Na) completely reacts with 60g of Chlorine (Cl2), what is the mass of the products?
a
40 g
b
60 g
c
100 g
d
160 g
The balanced equation for the reaction between sodium and chlorine is:
2Na + Cl2 -> 2NaCl
This means that for every 2 moles of Na, 2 moles of NaCl are produced.
The molar mass of Na is 22.99 g/mol, and the molar mass of Cl2 is 70.906 g/mol.
First, calculate the moles of Na and Cl2 in the given amounts:
moles of Na = (mass of Na) / (molar mass of Na)
moles of Na = 40 g / 22.99 g/mol
moles of Na = 1.74 mol
moles of Cl2 = (mass of Cl2) / (molar mass of Cl2)
moles of Cl2 = 60 g / 70.906 g/mol
moles of Cl2 = 0.846 mol
Since the reaction is a 1:1 ratio between Na and Cl2, we can only react the lesser amount of moles. Therefore, we can only react 0.846 mol of Na, which is equivalent to 0.846 mol of NaCl.
To calculate the mass of the products, determine the molar mass of NaCl and multiply it by the moles of NaCl:
molar mass of NaCl = 22.99 g/mol (molar mass of Na) + 35.45 g/mol (molar mass of Cl)
mass of NaCl = moles of NaCl * molar mass of NaCl
mass of NaCl = 0.846 mol * (22.99 g/mol + 35.45 g/mol)
mass of NaCl = 0.846 mol * 58.44 g/mol
mass of NaCl = 49.39 g
Therefore, the mass of the products (NaCl) is 49.39 g.
The correct answer is therefore:
c) 100 g
To find the mass of the products resulting from the reaction between sodium (Na) and chlorine (Cl2), we need to determine the limiting reactant, which is the reactant that is completely consumed and determines the amount of product formed.
First, we need to find the number of moles for each reactant. To do this, we divide the given masses by their respective molar masses.
The molar mass of sodium (Na) is 22.99 g/mol, so the number of moles of Na is calculated as:
Number of moles of Na = Mass of Na / Molar mass of Na
= 40 g / 22.99 g/mol
= 1.74 mol
The molar mass of chlorine (Cl2) is 70.90 g/mol, so the number of moles of Cl2 is calculated as:
Number of moles of Cl2 = Mass of Cl2 / Molar mass of Cl2
= 60 g / 70.90 g/mol
= 0.846 mol
Now we can determine the limiting reactant by comparing the moles of both reactants. The limiting reactant is the one with fewer moles. In this case, chlorine (Cl2) has fewer moles, so it is the limiting reactant.
Since the balanced chemical equation for the reaction between Na and Cl2 is:
2Na + Cl2 -> 2NaCl
We can see that the stoichiometric ratio between Na and Cl2 is 2:1. Therefore, for every 2 moles of Na consumed, 1 mole of Cl2 is consumed.
Since we have determined that chlorine (Cl2) is the limiting reactant, we need to calculate the moles of NaCl which can be formed. This can be done by multiplying the number of moles of Cl2 by the stoichiometric ratio:
Moles of NaCl = Moles of Cl2 * (2 moles of NaCl / 1 mole of Cl2)
= 0.846 mol * (2 / 1)
= 1.692 mol
Finally, we can calculate the mass of the products (NaCl) by multiplying the moles of NaCl by its molar mass, which is 58.44 g/mol:
Mass of NaCl = Moles of NaCl * Molar mass of NaCl
= 1.692 mol * 58.44 g/mol
= 98.82 g
Therefore, the mass of the products (NaCl) is approximately 98.82 grams.
The correct answer is (c) 100 g, which is the closest option available.
To determine the mass of the products, we need to start by finding the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
1. First, we convert the given masses of sodium (Na) and chlorine (Cl2) to moles using their molar masses. The molar mass of sodium (Na) is 22.99 g/mol, and the molar mass of chlorine (Cl2) is 35.45 g/mol.
Mass of Na = 40 g
Moles of Na = 40 g / 22.99 g/mol ≈ 1.74 mol
Mass of Cl2 = 60 g
Moles of Cl2 = 60 g / 35.45 g/mol ≈ 1.69 mol
2. Next, we determine the stoichiometry of the reaction between sodium and chlorine. From the balanced chemical equation for the reaction between sodium and chlorine:
2 Na + Cl2 → 2 NaCl
We can see that 2 moles of sodium (Na) react with 1 mole of chlorine (Cl2) to form 2 moles of sodium chloride (NaCl).
3. Now, we compare the moles of sodium (Na) and chlorine (Cl2) to determine the limiting reactant. Since the stoichiometry of the reaction is 2:1 for Na:Cl2, we need twice as many moles of chlorine as sodium for complete reaction.
Moles of Cl2 available / stoichiometric ratio = 1.69 mol / 1 = 1.69 mol
Moles of Na required / stoichiometric ratio = 1.74 mol / 2 = 0.87 mol
Since 0.87 mol of Na is less than 1.69 mol of Cl2, we can conclude that sodium (Na) is the limiting reactant.
4. Using the limiting reactant, we can determine the moles of product formed. Since 2 moles of NaCl are formed for every 2 moles of Na, and we have 0.87 mol of Na, we can calculate the moles of NaCl produced.
Moles of NaCl = Moles of Na (limiting reactant)
Moles of NaCl = 0.87 mol
5. Finally, we calculate the mass of NaCl formed using its molar mass, which is 58.44 g/mol.
Mass of NaCl = Moles of NaCl × Molar mass of NaCl
Mass of NaCl = 0.87 mol × 58.44 g/mol ≈ 50.76 g
Therefore, the mass of the products, specifically sodium chloride (NaCl), in this reaction is approximately 50.76 grams.