In the following chemical reaction between H2 and Cl2 to produce HCl, what is the mass of HCl produced with 0.35g of H2 completely reacts with 12.42g of Cl2?
12.77
12
To find the mass of HCl produced, we first need to determine the limiting reactant.
Step 1: Calculate the moles of each reactant
Molar mass of H2 = 2.02 g/mol
Molar mass of Cl2 = 70.90 g/mol
Moles of H2 = mass of H2 / molar mass of H2
= 0.35 g / 2.02 g/mol
≈ 0.173 mol
Moles of Cl2 = mass of Cl2 / molar mass of Cl2
= 12.42 g / 70.90 g/mol
≈ 0.175 mol
Step 2: Determine the limiting reactant
The limiting reactant is the one that is completely consumed in the reaction, determining the maximum amount of product that can be formed.
From the above calculation, we see that H2 and Cl2 are present in a 1:1 ratio, meaning that for every 1 mol of H2, 1 mol of Cl2 is required.
Since the number of moles of H2 (0.173 mol) is less than the number of moles of Cl2 (0.175 mol), H2 is the limiting reactant in this case.
Step 3: Calculate the mass of HCl produced
From the balanced chemical equation:
H2 + Cl2 → 2HCl
The stoichiometry of the reaction tells us that 1 mol of H2 produces 2 mol of HCl.
To find the mass of HCl produced, we can use the following equation:
Mass of HCl = (molar mass of HCl) x (moles of HCl)
The molar mass of HCl = 36.46 g/mol
Using the stoichiometry of the reaction, we can determine that the moles of HCl will be twice the moles of H2.
Moles of HCl = 2 x moles of H2
= 2 x 0.173 mol
= 0.346 mol
Now, we can calculate the mass of HCl produced:
Mass of HCl = (molar mass of HCl) x (moles of HCl)
= 36.46 g/mol x 0.346 mol
≈ 12.61 g
Therefore, the mass of HCl produced with 0.35 g of H2 completely reacting with 12.42 g of Cl2 is approximately 12.61 g.
To determine the mass of HCl produced in the given chemical reaction, we need to use stoichiometry, which relates the number of moles of reactants and products.
Step 1: Write and balance the chemical equation:
H2 + Cl2 -> 2HCl
Step 2: Calculate the number of moles of H2 and Cl2:
Given mass of H2 = 0.35g
Molar mass of H2 = 2.02 g/mol
Number of moles of H2 = mass / molar mass = 0.35g / 2.02 g/mol = 0.173 mol
Given mass of Cl2 = 12.42g
Molar mass of Cl2 = 70.90 g/mol
Number of moles of Cl2 = mass / molar mass = 12.42g / 70.90 g/mol = 0.175 mol
Step 3: Determine the limiting reactant:
To identify the limiting reactant, we compare the moles of each reactant based on their stoichiometric ratio in the balanced equation. The reactant with fewer moles is the limiting reactant.
From the balanced equation, we see that 1 mole of H2 reacts with 1 mole of Cl2 to produce 2 moles of HCl. Therefore, the ratio of moles of H2 to Cl2 is 1:1.
Since both reactants have nearly the same number of moles (0.173 mol for H2 and 0.175 mol for Cl2), they are in the same proportion. Hence, H2 is the limiting reactant.
Step 4: Calculate the moles of HCl produced:
Given that 1 mole of H2 reacts to produce 2 moles of HCl, we can use the stoichiometric ratio to calculate the moles of HCl produced.
Moles of H2 = 0.173 mol
Moles of HCl produced = 2 moles of HCl per 1 mole of H2 * 0.173 mol of H2 = 0.346 mol
Step 5: Calculate the mass of HCl produced:
To find the mass of HCl produced, we use the molar mass of HCl.
Molar mass of HCl = 36.46 g/mol
Mass of HCl produced = moles of HCl * molar mass of HCl = 0.346 mol * 36.46 g/mol = 12.63 g
Therefore, the mass of HCl produced when 0.35 g of H2 completely reacts with 12.42 g of Cl2 is approximately 12.63 g.