Calculate the volume of amonia gas is produced at STP when 140g of nitrogen gas reacts with 30g of hydrogen gas (Atomic mass N=14u,H=1u)
N2 + 3H2 -> 2NH3
140g N2 = 5 moles
30g H2 = 15 moles
So, you will get 10 moles NH3
1 mole occupies 22.4 L, so ...
To calculate the volume of ammonia gas produced at Standard Temperature and Pressure (STP), we need to use stoichiometry and the ideal gas law.
1. Write the balanced chemical equation for the reaction:
N2 + 3H2 -> 2NH3
2. Convert the mass of nitrogen (N2) to moles using the molar mass of nitrogen:
Moles of N2 = Mass of N2 / Molar mass of N2
= 140g / 28g/mol
= 5 moles
3. Convert the mass of hydrogen (H2) to moles using the molar mass of hydrogen:
Moles of H2 = Mass of H2 / Molar mass of H2
= 30g / 2g/mol
= 15 moles
4. Determine the limiting reactant:
The balanced equation tells us that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
So, for 5 moles of N2, we need 15 moles of H2.
Since we have 15 moles of H2 available, H2 is in excess and N2 is the limiting reactant.
5. Calculate the moles of NH3 produced:
From the balanced equation, we know that 1 mole of N2 produces 2 moles of NH3.
Moles of NH3 = Moles of N2 x (2 moles of NH3 / 1 mole of N2)
= 5 moles x (2/1)
= 10 moles
6. Now, we can use the ideal gas law to calculate the volume of ammonia gas at STP.
The ideal gas law equation is: PV = nRT
P = pressure = 1 atm (STP)
V = volume of ammonia gas (to be calculated)
n = moles of NH3 = 10 moles
R = ideal gas constant = 0.0821 L.atm/mol.K (at STP)
T = temperature = 273.15 K (at STP)
Rearranging the ideal gas law equation, we get:
V = nRT / P = (10 moles)(0.0821 L.atm/mol.K)(273.15 K) / 1 atm
= 224.2 liters
Therefore, the volume of ammonia gas produced at STP when 140g of nitrogen gas reacts with 30g of hydrogen gas is 224.2 liters.
To calculate the volume of ammonia gas produced at STP (Standard Temperature and Pressure), we need to use the balanced chemical equation and the stoichiometry of the reaction.
The balanced chemical equation for the reaction between nitrogen gas (N2) and hydrogen gas (H2) to form ammonia gas (NH3) is:
N2 + 3H2 → 2NH3
First, we need to determine the number of moles of nitrogen gas and hydrogen gas that we have using their respective masses and molar masses.
Molar mass of N2 = 2 × atomic mass of N = 2 × 14 g/mol = 28 g/mol
Molar mass of H2 = 2 × atomic mass of H = 2 × 1 g/mol = 2 g/mol
Number of moles of N2 = mass of N2 / molar mass of N2 = 140 g / 28 g/mol = 5 mol
Number of moles of H2 = mass of H2 / molar mass of H2 = 30 g / 2 g/mol = 15 mol
From the balanced chemical equation, we can see that 1 mole of nitrogen gas reacts with 3 moles of hydrogen gas to produce 2 moles of ammonia gas.
Using the stoichiometry of the reaction, we can determine the limiting reactant - the reactant that is completely consumed and determines the amount of product formed. In this case, nitrogen gas is the limiting reactant because for every 1 mole of nitrogen gas, 3 moles of hydrogen gas are required. Since we have 5 moles of nitrogen gas and only 15 moles of hydrogen gas, we have enough hydrogen gas to fully react with the nitrogen gas.
Therefore, 5 moles of nitrogen gas will react to produce 2 moles of ammonia gas.
The volume of a gas, at STP, is directly proportional to the number of moles of the gas. At STP, one mole of any ideal gas occupies 22.4 liters of volume.
So, the volume of ammonia gas produced at STP can be calculated as:
Volume of ammonia gas = (Number of moles of ammonia gas) × (22.4 L/mol)
Number of moles of ammonia gas = Number of moles of nitrogen gas (because of the stoichiometry of the reaction)
Number of moles of ammonia gas = 5 mol
Volume of ammonia gas = 5 mol × 22.4 L/mol = 112 L
Therefore, the volume of ammonia gas produced at STP when 140g of nitrogen gas reacts with 30g of hydrogen gas is 112 liters.