Calculate the mass of ammonia produced when 11.2cm3 of nitrogen gas is reacted with excess hydrogen gas
To calculate the mass of ammonia produced when 11.2 cm³ of nitrogen gas is reacted with excess hydrogen gas, we need to determine the balanced chemical equation for the reaction and use stoichiometry.
The balanced chemical equation for the reaction between nitrogen gas (N₂) and hydrogen gas (H₂) to form ammonia (NH₃) is:
N₂ + 3H₂ -> 2NH₃
From the balanced equation, we can see that 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃.
To find the moles of N₂, we need to convert the given volume (11.2 cm³) to moles using the ideal gas law equation:
n = PV / RT
where:
n = moles of gas
P = pressure of gas (assumed constant)
V = volume of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (assumed constant)
Next, we'll need to convert moles of N₂ to moles of NH₃ using the stoichiometry of the balanced equation.
Finally, we can convert moles of NH₃ to grams using the molar mass of NH₃, which is 17.03 g/mol.
Let's calculate the mass of ammonia produced step-by-step:
Step 1: Convert volume of nitrogen gas to moles
Using the ideal gas law:
n(N₂) = PV / RT
Assuming standard temperature and pressure (STP), P and T can be taken as 1 atm and 273 K (0 °C), respectively.
Let's assume the pressure and temperature are constant.
n(N₂) = (1 atm) * (11.2 cm³) / (0.0821 L·atm/mol·K * 273 K)
Step 2: Convert moles of N₂ to moles of NH₃ using the stoichiometry of the balanced equation.
From the balanced equation, we know that 1 mole of N₂ produces 2 moles of NH₃. So we multiply the moles of N₂ by 2:
n(NH₃) = 2 * n(N₂)
Step 3: Convert moles of NH₃ to grams.
Molar mass of NH₃ = 17.03 g/mol
Mass(MNH₃) = n(NH₃) * (molar mass of NH₃)
Let's calculate the mass of ammonia produced using these steps:
Step 1:
n(N₂) = (1 atm) * (11.2 cm³) / (0.0821 L·atm/mol·K * 273 K) = 0.0004999 mol of N₂
Step 2:
n(NH₃) = 2 * 0.0004999 mol of N₂ = 0.0009998 mol of NH₃
Step 3:
Mass(MNH₃) = 0.0009998 mol of NH₃ * 17.03 g/mol = 0.01705 g
Therefore, the mass of ammonia produced when 11.2 cm³ of nitrogen gas is reacted with excess hydrogen gas is approximately 0.01705 grams.
To calculate the mass of ammonia produced, we need to use stoichiometry - the relationship between the balanced equation and the amounts of substances involved. In this case, we need to know the balanced equation for the reaction between nitrogen gas (N2) and hydrogen gas (H2) to form ammonia (NH3).
The balanced equation for this reaction is:
N2 + 3H2 -> 2NH3
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 (NH3).
Now, let's convert the given volume of nitrogen gas to moles. To do this, we need to know the conditions under which the volume is given (temperature and pressure).
Once we have the moles of nitrogen gas, we can use the stoichiometry of the balanced equation to calculate the moles of ammonia produced.
Finally, to calculate the mass of ammonia produced, multiply the moles of ammonia by its molar mass. The molar mass of ammonia (NH3) is approximately 17.03 g/mol.
Without the given conditions, it is not possible to provide an accurate calculation. Please provide the temperature and pressure at which the given volume of nitrogen gas (11.2 cm3) is measured, and I will be able to guide you through the calculation.
N2 + 3H2 ==> 2NH3
When gases are used one may use cc directly as if they were mols.
11.2 cc N2 x (2 mols NH3/1 mol N2) = 11.2 x 2/1 = 22.4 cc NH3 gas produced.