How much 0.5 M HCl is needed to neutralize 5.5 kg of ammonia dissolved in 6 gal of water?
NH3 + HCl ==> NH4Cl
5.5 kg = 5500 g
Convert to moles.
moles = grams/molar mass = ?
1 mole NH3 requires 1 mol HCl; therefore, moles NH3 = moles HCl.
M HCl = moles HCl/L HCl.
You know M HCl and moles HCl, solve for L HCl
To determine the amount of 0.5 M HCl needed to neutralize ammonia, we need to calculate the amount of moles of ammonia and then use the stoichiometry of the neutralization reaction.
1. Calculate the amount of ammonia in moles:
To do this, we need to know the molar mass of ammonia (NH3). The molar mass of NH3 is approximately 17 g/mol.
Given:
- Mass of ammonia = 5.5 kg = 5500 g
Using the molar mass, we can determine the amount of ammonia in moles:
moles of NH3 = mass / molar mass = 5500 g / 17 g/mol ≈ 323.53 mol
2. Determine the stoichiometry of the reaction:
The neutralization reaction between ammonia (NH3) and hydrochloric acid (HCl) can be represented by the following balanced equation:
NH3 + HCl → NH4Cl
According to the equation, one mole of HCl reacts with one mole of NH3.
3. Calculate the amount of 0.5 M HCl required to neutralize the ammonia:
Using the molarity (concentration) of HCl and the stoichiometry of the reaction, we can determine the amount of HCl needed.
Given:
- Molarity of HCl = 0.5 M
moles of HCl = moles of NH3 = 323.53 mol
The volume of HCl required can be calculated using the formula:
Volume (in L) = moles / molarity
volume of HCl = 323.53 mol / 0.5 mol/L = 647.06 L
However, we also need to consider the volume of water in the solution.
Given:
- Volume of water = 6 gal
To convert gallons to liters, we use the conversion factor: 1 gal = 3.78541 L
volume of water = 6 gal × 3.78541 L/gal = 22.71246 L
Therefore, the total volume of the solution is:
total volume = volume of HCl + volume of water = 647.06 L + 22.71246 L = 669.77246 L
So, approximately 669.77 liters of 0.5 M HCl is needed to neutralize 5.5 kg of ammonia dissolved in 6 gal of water.
To determine how much 0.5 M HCl is needed to neutralize the ammonia solution, we need to calculate the moles of ammonia present in the solution and then find the corresponding moles of HCl required for neutralization.
Step 1: Convert the mass of ammonia to moles.
Given that the ammonia is dissolved in 6 gallons of water, we need to calculate the mass of pure ammonia in the solution. The density of water is approximately 1 kg/L or 8.34 lb/gal, so we can convert the 6 gallons to kilograms:
6 gallons * 8.34 lb/gal * (1 kg/2.2046 lb) = 2.73 kg
So, the mass of ammonia in the solution is 5.5 kg - 2.73 kg = 2.77 kg.
Now, we'll convert the mass of ammonia to moles using its molar mass. The molar mass of ammonia (NH3) is approximately 17.03 g/mol.
2.77 kg * (1000 g/kg) / (17.03 g/mol) = 162.6 mol
Step 2: Calculate the moles of HCl required for neutralization.
The balanced chemical equation for the neutralization reaction between HCl and NH3 is:
HCl + NH3 → NH4Cl
The stoichiometric ratio between HCl and NH3 is 1:1, meaning that each mole of HCl reacts with one mole of NH3.
Therefore, the moles of HCl required for neutralization is equal to the moles of ammonia in the solution, which we calculated to be 162.6 mol.
Step 3: Convert moles of HCl to volume using the molarity.
The molarity (M) of HCl is 0.5 M, meaning that it contains 0.5 mol of HCl per liter of solution.
To calculate the volume of 0.5 M HCl required, we divide the moles of HCl by the molarity:
Volume of HCl = Moles of HCl / Molarity
Volume of HCl = 162.6 mol / 0.5 mol/L = 325.2 L
However, our initial given volume was in gallons, so we need to convert liters back to gallons:
325.2 L * (1 gal / 3.7854 L) = 85.9 gal
Therefore, approximately 85.9 gallons of 0.5 M HCl is needed to neutralize 5.5 kg of ammonia dissolved in 6 gallons of water.