How many grams of NH3 are present in 725 mL of gas at STP?
1 mol NH3 (17 grams) is present in a volume of 22.4 L at STP. You have 0.725L
To determine the number of grams of NH3 present in 725 mL of gas at STP, we can use the ideal gas law equation:
PV = nRT
Where:
P = Pressure (STP = 1 atm)
V = Volume (725 mL = 0.725 L)
n = number of moles
R = Ideal gas constant (0.0821 L.atm/mol.K)
T = Temperature (STP = 273.15 K)
First, let's calculate the number of moles:
n = (PV) / (RT)
= (1 atm * 0.725 L) / (0.0821 L.atm/mol.K * 273.15 K)
Calculating this expression:
n ≈ 0.0278 moles
Next, we need to convert moles to grams using the molar mass of NH3:
Molar mass of NH3 = 14.01 g/mol (N) + 3(1.01 g/mol) (H)
= 17.03 g/mol
Finally, we can calculate the grams of NH3:
Grams = n * molar mass
≈ 0.0278 moles * 17.03 g/mol
Calculating this expression:
Grams ≈ 0.473 g
Therefore, there are approximately 0.473 grams of NH3 present in 725 mL of gas at STP.
To find the number of grams of NH3 present in 725 mL of gas at STP (Standard Temperature and Pressure), we can use the ideal gas law equation and the molar mass of NH3.
First, let's determine the number of moles of NH3 using the ideal gas law equation:
PV = nRT
Where:
P = pressure (at STP, it is 1 atm)
V = volume (725 mL, which we need to convert to liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (at STP, it is 273.15 K)
Converting the volume to liters:
725 mL * (1 L / 1000 mL) = 0.725 L
Now, let's substitute the values into the ideal gas law equation:
(1 atm) * (0.725 L) = n * (0.0821 L·atm/(mol·K)) * (273.15 K)
Simplifying the equation:
0.725 = n * 22.414
n ≈ 0.0324 mol
Next, we need to determine the molar mass of NH3 (ammonia). The molar mass of nitrogen (N) is 14.01 g/mol, and the molar mass of hydrogen (H) is 1.008 g/mol. Ammonia (NH3) consists of one nitrogen atom and three hydrogen atoms.
Molar mass of NH3:
(1 × 14.01 g/mol) + (3 × 1.008 g/mol) = 17.03 g/mol
Now, let's calculate the mass of NH3 using the number of moles and molar mass:
Mass of NH3 = number of moles (n) * molar mass
Mass of NH3 ≈ 0.0324 mol * 17.03 g/mol
Mass of NH3 ≈ 0.552 g
Therefore, there are approximately 0.552 grams of NH3 present in 725 mL of gas at STP.