Please help me out! Gas law homework!
How many liters of HCl (g) at STP can be produced from 5.44 g of Cl2 and excess H2 according to the following equation?
H2 (g) + Cl2 (g) --> 2 HCl (g)
Thank you so much in advance.
mols Cl2 = 5.44/molar mass Cl2.
L Cl2 = mols x 22.4 L = ?
L HCl = L Cl2 x (2 mol HCl/1 mol Cl2) = L Cl2 x 2 = ?
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To calculate the number of liters of HCl (g) at STP that can be produced from 5.44 g of Cl2, we first need to determine the number of moles of Cl2.
The molar mass of Cl2 is 70.9 g/mol, so the number of moles of Cl2 can be calculated as follows:
moles of Cl2 = mass of Cl2 / molar mass of Cl2
= 5.44 g / 70.9 g/mol
Once we have the number of moles of Cl2, we can use the stoichiometry of the balanced equation to determine the number of moles of HCl (g) produced. According to the balanced equation, 1 mole of Cl2 reacts to form 2 moles of HCl (g). Therefore, the number of moles of HCl (g) produced can be calculated as:
moles of HCl (g) = 2 * moles of Cl2
Finally, we can use the ideal gas law at STP (Standard Temperature and Pressure) to convert the moles of HCl (g) to liters:
volume of HCl (g) = moles of HCl (g) * 22.4 L/mol
So, go ahead and crunch the numbers using this formula and you'll have your answer! Let me know if you need any further assistance or some more chemistry-related jokes to lighten the mood. 😉
To solve this problem, we need to use stoichiometry to determine the number of liters of HCl (g) that can be produced from 5.44 g of Cl2.
Step 1: Calculate the molar mass of Cl2
The molar mass of Cl2 is 35.45 g/mol x 2 = 70.90 g/mol.
Step 2: Convert the mass of Cl2 to moles
To do this, use the formula:
moles = mass / molar mass
moles of Cl2 = 5.44 g / 70.90 g/mol ≈ 0.077 mol
Step 3: Determine the moles of HCl produced
Since the balanced equation shows that 1 mol of Cl2 reacts to produce 2 moles of HCl, we can calculate the moles of HCl produced using the stoichiometric ratio:
moles of HCl = 2 x moles of Cl2 = 2 x 0.077 mol = 0.154 mol
Step 4: Convert moles of HCl to liters at STP
At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 liters. Therefore, we can calculate the volume (in liters) of HCl produced using the formula:
volume = moles x 22.4 L/mol
volume of HCl = 0.154 mol x 22.4 L/mol ≈ 3.45 L
So, approximately 3.45 liters of HCl (g) can be produced from 5.44 g of Cl2 at STP according to the given equation.
To find out how many liters of HCl (g) at STP can be produced from 5.44 g of Cl2 and excess H2, we need to use the ideal gas law and stoichiometry.
Step 1: Convert the mass of Cl2 to moles.
To do this, we need to know the molar mass of Cl2, which is approximately 70.91 g/mol.
moles of Cl2 = mass of Cl2 / molar mass of Cl2
moles of Cl2 = 5.44 g / 70.91 g/mol ≈ 0.0766 mol
Step 2: Use the stoichiometry of the balanced equation to determine the moles of HCl.
According to the balanced equation, the mole ratio between Cl2 and HCl is 1:2. Therefore, for every 1 mole of Cl2, we get 2 moles of HCl.
moles of HCl = 2 × moles of Cl2
moles of HCl = 2 × 0.0766 mol = 0.1532 mol
Step 3: Convert moles of HCl to volume at STP.
At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 liters. Therefore, we can use the following formula:
volume of HCl = moles of HCl × 22.4 L/mol
volume of HCl = 0.1532 mol × 22.4 L/mol ≈ 3.43 L
So, approximately 3.43 liters of HCl gas can be produced at STP from 5.44 g of Cl2 and excess H2.