0.84grams or aluminum reacted completely with chlorine gas . calculate the volume of chlorine
To calculate the volume of chlorine gas that reacted with 0.84 grams of aluminum, we need to use the stoichiometry of the reaction between aluminum and chlorine.
The balanced chemical equation for the reaction is:
2Al + 3Cl2 -> 2AlCl3
This means that 2 moles of aluminum react with 3 moles of chlorine gas to produce 2 moles of aluminum chloride.
First, we need to find the number of moles of aluminum in 0.84 grams. The molar mass of aluminum is 26.98 g/mol, so:
0.84 g Al / 26.98 g/mol Al = 0.031 moles Al
According to the stoichiometry of the balanced equation, 2 moles of aluminum react with 3 moles of Cl2. Therefore, we can use a proportion to calculate the number of moles of Cl2 that react with 0.031 moles of Al:
2 moles Al : 3 moles Cl2 = 0.031 moles Al : x moles Cl2
x = (0.031 moles Al)(3 moles Cl2 / 2 moles Al) = 0.047 moles Cl2
Finally, we can use the ideal gas law to calculate the volume of chlorine gas that corresponds to 0.047 moles at standard temperature and pressure (STP).
PV = nRT
where P = 1 atm (STP pressure), V is the volume of chlorine gas, n = 0.047 moles, R = 0.082 L·atm/mol·K (gas constant), and T = 273 K (STP temperature).
Solving for V:
V = nRT / P = (0.047 mol)(0.082 L·atm/mol·K)(273 K) / (1 atm) ≈ 1.1 L
Therefore, the volume of chlorine gas that reacted with 0.84 grams of aluminum is approximately 1.1 liters.
To determine the volume of chlorine gas (Cl2) that reacted with 0.84 grams of aluminum (Al), we need to use the stoichiometry of the balanced chemical equation.
The balanced chemical equation for this reaction is:
2 Al + 3 Cl2 -> 2 AlCl3
From the balanced equation, we can see that 2 moles of aluminum react with 3 moles of chlorine (Cl2) to form 2 moles of aluminum chloride (AlCl3).
Step 1: Calculate the number of moles of aluminum (Al) in 0.84 grams:
Molar mass of aluminum (Al) = 26.98 g/mol
Number of moles of aluminum (Al) = Mass / Molar mass
Number of moles of aluminum (Al) = 0.84 g / 26.98 g/mol
Step 2: Convert the moles of aluminum (Al) to moles of chlorine (Cl2) using the stoichiometry of the balanced equation.
From the balanced equation, we know that 2 moles of Al react with 3 moles of Cl2.
Therefore, Moles of Cl2 = (Moles of Al) x (3 moles of Cl2 / 2 moles Al)
Step 3: Convert the moles of chlorine (Cl2) to volume:
1 mole of any gas at standard temperature and pressure (STP) occupies 22.4 liters.
Volume of Cl2 = (Moles of Cl2) x (22.4 liters/mole)
Let's plug in the values:
Moles of Al = 0.84 g / 26.98 g/mol
Moles of Cl2 = (Moles of Al) x (3 moles Cl2 / 2 moles Al)
Volume of Cl2 = (Moles of Cl2) x (22.4 liters/mole)
Calculate each step accordingly to find the volume of chlorine gas that reacted with 0.84 grams of aluminum.
To calculate the volume of chlorine gas, we need to use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.
First, let's calculate the number of moles of aluminum using its molar mass. The molar mass of aluminum is 26.98 g/mol.
Moles of aluminum = Mass of aluminum / Molar mass of aluminum
= 0.84 g / 26.98 g/mol
= 0.03118 mol
Next, we need to determine the stoichiometry of the reaction to find the number of moles of chlorine gas produced. The balanced chemical equation for this reaction is:
2Al + 3Cl₂ → 2AlCl₃
According to the equation, 2 moles of aluminum react with 3 moles of chlorine gas to form 2 moles of aluminum chloride.
Since the stoichiometry shows that 2 moles of aluminum react to produce 3 moles of chlorine gas, we can set up a ratio:
Moles of chlorine gas / Moles of aluminum = 3 / 2
Plugging in the value for moles of aluminum (0.03118 mol):
Moles of chlorine gas = (0.03118 mol * 3) / 2
= 0.04677 mol
Now, we have the number of moles of chlorine gas. To calculate the volume, we need to know the temperature and pressure. Without that information, we cannot proceed with calculating the volume.