How many grams of Mg must react with HCl in order to produce 40.0 ml of hydrogen gas at 22 degrees celsius and 1.02 bar?
If the magnesium ribbon weighs 1.83mg/mm, how many mm of the ribbon will you require?
Thanks in advance for your help!!
Mg + 2HCl ==> H2 + MgCl2
Use PV = nRT and solve for n = mols H2 gas at the conditions of the problem.
Use the coefficients in the balanced equation to convert mols H2 to mols Mg.
Then g Mg = mols Mg x atomic mass Mg (in grams). Convert to mg.
Finally, length of Mg is
1.83 mg/mm x ?mm = mg Mg from the last step. Solve for ?mm.
To determine the amount of magnesium (Mg) needed to react with hydrochloric acid (HCl) and produce 40.0 ml of hydrogen gas at 22 degrees Celsius and 1.02 bar, we need to follow these steps:
Step 1: Identify the balanced chemical equation for the reaction between Mg and HCl.
The balanced equation for this reaction is: Mg + 2HCl → MgCl2 + H2
From the equation, we can see that 1 mole of Mg reacts with 2 moles of HCl to produce 1 mole of H2.
Step 2: Calculate the number of moles of H2 gas.
We have the volume of the H2 gas, which is 40.0 ml. To convert this volume into moles, we need to use the ideal gas law: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin.
First, we need to convert the given temperature from Celsius to Kelvin by adding 273.15:
T = 22 + 273.15 = 295.15 K
Now, we can rearrange the ideal gas law equation to solve for n:
n = PV / RT
n = (1.02 bar) * (40.0 ml) / (0.0821 L atm/mol K * 295.15 K)
Note that we converted the volume from ml to L by dividing by 1000.
n = 0.0016578 mol
Step 3: Calculate the number of moles of Mg needed.
From the balanced equation, we know that 1 mole of Mg produces 1 mole of H2. Therefore, we need the same number of moles of Mg as the number of moles of H2.
Step 4: Convert moles of Mg to grams.
To convert moles of Mg to grams, we need to know the molar mass of Mg, which is approximately 24.31 g/mol.
Mass of Mg = number of moles of Mg * molar mass of Mg
Mass of Mg = 0.0016578 mol * 24.31 g/mol
Therefore, the mass of Mg required is 0.0402 grams.
Now, we need to find out how many millimeters (mm) of the magnesium ribbon will be required.
We are given that the magnesium ribbon weighs 1.83 mg/mm. So to find the length of the ribbon required, we need to use the following steps:
Step 1: Convert the mass of Mg calculated (0.0402 grams) to milligrams (mg).
Mass of Mg in mg = 0.0402 g * 1000 = 40.2 mg
Step 2: Use the given weight of the ribbon in mg per mm to calculate the length of ribbon required.
Length of ribbon = Mass of Mg in mg / weight of the ribbon in mg per mm
Length of ribbon = 40.2 mg / 1.83 mg/mm
Therefore, the length of the magnesium ribbon required is approximately 21.97 mm (rounded to the nearest hundredth).
Remember to always double-check your calculations and consider any uncertainties or rounding needed for your final answer.
To find out how many grams of Mg are needed to produce 40.0 ml of hydrogen gas, we need to use the stoichiometry of the reaction between Mg and HCl.
Step 1: Write the balanced chemical equation for the reaction:
Mg + 2HCl -> MgCl2 + H2
Step 2: Find the molar volume of hydrogen gas at STP (standard temperature and pressure).
At STP, 1 mole of any ideal gas occupies 22.4 L. Therefore, the molar volume of hydrogen gas is 22.4 L/mol.
Step 3: Convert the given volume of hydrogen gas to liters.
40.0 ml is equal to 40.0/1000 = 0.04 L.
Step 4: Use the molar volume at STP to find the number of moles of hydrogen gas produced.
The number of moles of H2 = volume of H2 gas / molar volume of H2 gas
= 0.04 L / 22.4 L/mol
Step 5: Use the stoichiometry of the balanced equation to find the number of moles of Mg required.
From the balanced equation, we know that 1 mole of Mg reacts with 1 mole of H2. Therefore, the moles of Mg required would be equal to the moles of H2 produced.
Step 6: Convert the moles of Mg to grams using the molar mass of Mg.
The molar mass of Mg is 24.31 g/mol.
Step 7: Calculate the grams of Mg required.
Grams of Mg = Moles of Mg x Molar mass of Mg
Now, let's solve the problem step-by-step.
Step 1: The balanced chemical equation for the reaction is:
Mg + 2HCl -> MgCl2 + H2
Step 2: The molar volume of hydrogen gas at STP is 22.4 L/mol.
Step 3: The given volume of hydrogen gas is 0.04 L.
Step 4: The number of moles of hydrogen gas produced is:
0.04 L / 22.4 L/mol = 0.00179 mol
Step 5: The number of moles of Mg required is also 0.00179 mol.
Step 6: The molar mass of Mg is 24.31 g/mol.
Step 7: The grams of Mg required is:
0.00179 mol x 24.31 g/mol = 0.0436 g
Therefore, you will require 0.0436 grams of Mg to produce 40.0 ml of hydrogen gas.
Additionally, the mass of the magnesium ribbon required can be determined using the given information.
Step 8: The mass of the magnesium ribbon is given as 1.83 mg/mm.
Step 9: Convert the mass of Mg required from grams to milligrams.
0.0436 g x 1000 mg/g = 43.6 mg
Step 10: Use the mass per unit length of the ribbon to find the length required.
Length of Mg ribbon required = mass of Mg required / mass per unit length of the ribbon
= 43.6 mg / 1.83 mg/mm
Step 11: Calculate the length of the magnesium ribbon required.
Length of Mg ribbon required = 23.8 mm
Therefore, you will require 23.8 mm of the magnesium ribbon to react with HCl and produce 40.0 ml of hydrogen gas at 22 degrees Celsius and 1.02 bar.