N2(g)+H2(g)-NH3(g)
Calculate the volume of ammonia gas produced when20cm3 of nitrogen gas was reacted with an excess of hydrogen gas
N2 + 3H2 ==> 2NH3
Is that at STP?
20 cc x (1mol/22,400) = mols N2.
Convert mols N2 to mols NH3 using the coefficients in the balanced equation.
Now convert mols NH3 to L. mol x 22.4L/mol = L
6.72lit
To calculate the volume of ammonia gas (NH3) produced, we will use the balanced chemical equation for the reaction:
N2(g) + 3H2(g) → 2NH3(g)
From the balanced equation, we can see that 1 mole of nitrogen gas (N2) reacts with 3 moles of hydrogen gas (H2) to produce 2 moles of ammonia gas (NH3).
Step 1: Convert the given volume of nitrogen gas (N2) from cm3 to moles.
Given: Volume of N2 = 20 cm3
To convert cm3 to moles, we need to use the ideal gas equation, PV = nRT, where:
P is the pressure (which is not given)
V is the volume in liters
n is the number of moles
R is the ideal gas constant (0.0821 L·atm/mol·K)
T is the temperature (which is not given)
Since the pressure and temperature are not provided, we can assume they are at standard conditions (STP), which is 1 atm and 273 K.
Using the ideal gas equation, we can calculate the number of moles as follows:
n = (V × P) / (R × T)
n = (20 cm3) / (1000 cm3/L) [converting cm3 to liters]
n = 0.02 L / (0.0821 L·atm/mol·K × 273 K) [using standard conditions]
n ≈ 0.00088 moles (rounded to 5 decimal places)
Step 2: Use the mole ratio from the balanced equation to determine the moles of ammonia gas produced.
According to the balanced equation, for every 1 mole of nitrogen gas reacted, 2 moles of ammonia gas are produced. Since we have calculated that there are 0.00088 moles of nitrogen gas, the moles of ammonia gas produced would be:
Moles of NH3 = (0.00088 moles of N2) × (2 moles of NH3 / 1 mole of N2)
Moles of NH3 ≈ 0.00176 moles (rounded to 5 decimal places)
Step 3: Convert the moles of ammonia gas to volume (in liters).
To convert moles to volume, we can use the same ideal gas equation as before:
V = (n × R × T) / P
V = (0.00176 moles) × (0.0821 L·atm/mol·K) × (273 K) / (1 atm)
V ≈ 0.0407 L (rounded to 4 decimal places)
Step 4: Convert the volume to cm3.
To convert liters to cm3, we need to multiply the volume by 1000:
Volume of NH3 = 0.0407 L × 1000 cm3/L
Volume of NH3 ≈ 40.7 cm3 (rounded to 1 decimal place)
Therefore, the volume of ammonia gas produced when 20 cm3 of nitrogen gas was reacted with an excess of hydrogen is approximately 40.7 cm3.