How many grams of NH3 is there in 800 cm^3 at STP of NH3 gas?
22.4 L NH3 at STP has a mass of 1 mol or 17 grams. So 800 cc (0.800 L) will contain
17 x (0.800/22.4) = ?
To determine the number of grams of NH3 in 800 cm^3 at STP, we need to use the ideal gas law and molar mass of NH3.
Step 1: Convert cm^3 to liters:
800 cm^3 = 800/1000 = 0.8 liters
Step 2: Use the ideal gas law equation:
PV = nRT
Where:
P = pressure (at STP, it is 1 atm)
V = volume (0.8 liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (at STP, it is 273.15 K)
Step 3: Solve for n (number of moles):
n = (PV) / (RT)
n = (1 atm * 0.8 L) / (0.0821 L·atm/mol·K * 273.15 K)
Step 4: Calculate the molar mass of NH3:
Molar mass of NH3 = (1 x 14.01 g/mol) + (3 x 1.01 g/mol)
Molar mass of NH3 = 17.03 g/mol
Step 5: Calculate the number of grams of NH3:
grams = n * molar mass of NH3
Now let's plug in the values and calculate the answer:
n = (1 atm * 0.8 L) / (0.0821 L·atm/mol·K * 273.15 K)
n = 0.02898 mol
grams = 0.02898 mol * 17.03 g/mol
grams ≈ 0.49357 g
Therefore, there are approximately 0.49357 grams of NH3 in 800 cm^3 at STP.
To find the number of grams of NH3 in 800 cm^3 at STP (Standard Temperature and Pressure) of NH3 gas, we need to use the ideal gas law. The ideal gas law is expressed as:
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
R is the ideal gas constant
T is the temperature of the gas in Kelvin
At STP, the values for pressure (P) and temperature (T) are defined as:
P = 1 atm
T = 273 K
The ideal gas constant (R) is approximately 0.0821 L•atm/(mol•K).
In this case, we are given the volume (V) of NH3 gas, which is 800 cm^3. However, to use the ideal gas law, we need to convert the volume from cm^3 to liters (L), since the ideal gas constant is in units of L.
1 L = 1000 cm^3
Converting the volume, we have:
V = 800 cm^3 = 800/1000 L = 0.8 L
Now, we can plug in the values into the ideal gas law equation to find the number of moles (n) of NH3 gas:
PV = nRT
(1 atm)(0.8 L) = n(0.0821 L•atm/(mol•K))(273 K)
Simplifying the equation:
0.8 atm•L = 0.0821 n mol•K
n = (0.8 atm•L) / (0.0821 mol•K)
Now, we can calculate the number of moles (n):
n = 9.7598 mol
To convert moles to grams, we need to know the molar mass of NH3. The molar mass of NH3 is approximately 17.03 g/mol, which is the sum of the atomic masses of nitrogen (N) and three hydrogen atoms (H):
Molar mass of NH3 = (1 * atomic mass of N) + (3 * atomic mass of H)
= (1 * 14.01 g/mol) + (3 * 1.01 g/mol)
= 17.03 g/mol
Finally, we can calculate the number of grams (m) of NH3:
m = n * molar mass of NH3
= 9.7598 mol * 17.03 g/mol
= 166.23 g
Therefore, there are approximately 166.23 grams of NH3 in 800 cm^3 at STP of NH3 gas.