In the reaction N2 + 3H2 -> 2NH3, how many grams of NH3 gas would be formed when H2 gas, having a volume of 11.2 liters at STP, reacts with excess N2 gas?
Completly confused, thank you very much for your help :)
2NH3 mass = 17*2 = 34 g
3H2 mass = 6 g
11.2 liters H2 = 11.2/22.4 = 0.5 mole = 0.5 * 2 = 1 g
34 x = 6 1
x = 5.667
5.7 grams NH3 is required for 11.2 liters H2.
edited: last one didn't load divison signs
2NH3 mass = 17*2 = 34 g
3H2 mass = 6 g
11.2 liters H2 = 11.2/22.4 = 0.5 mole = 0.5 * 2 = 1 g
34/x = 6/1
X = 5.667
5.7 grams NH3 is required for 11.2 liters H2.
To find the number of grams of NH3 gas formed in the given reaction, we need to use the concept of stoichiometry and the ideal gas law at STP (Standard Temperature and Pressure).
1. Start by determining the number of moles of H2 gas using the ideal gas law equation:
PV = nRT
At STP, the pressure (P) is 1 atmosphere (atm), the volume (V) is 11.2 liters, the gas constant (R) is 0.0821 L.atm/mol.K, and the temperature (T) is 273.15 Kelvin (K).
Rearrange the equation and solve for n (the number of moles):
n = PV / RT
n = (1 atm) * (11.2 L) / (0.0821 L.atm/mol.K * 273.15 K)
This gives you the number of moles of H2 gas.
2. The balanced chemical equation shows that 3 moles of H2 produce 2 moles of NH3. Therefore, we need to convert the moles of H2 to moles of NH3 using the stoichiometric ratio:
Moles of NH3 = Moles of H2 * (2 moles NH3 / 3 moles H2)
3. Next, we need to convert the moles of NH3 to grams by using the molar mass of NH3. The molar mass of NH3 is calculated by adding the atomic masses of nitrogen (N) and hydrogen (H). The atomic mass of nitrogen is 14.01 g/mol, and hydrogen is 1.01 g/mol.
Molar mass of NH3 = (14.01 g/mol) + (3 * 1.01 g/mol)
4. Finally, multiply the moles of NH3 by the molar mass of NH3 to get the grams of NH3:
Grams of NH3 = Moles of NH3 * Molar mass of NH3
By following these steps, you can calculate the grams of NH3 gas formed when H2 gas reacts with excess N2 gas at STP.
Given the following chemical reaction:
N2 + 3H2 → 2NH3
How many grams of NH3 would be produced if you started out with 56 grams of nitrogen gas and 10 grams of hydrogen gas?
Here is a link that will show you how to work stoichiometry problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html
There is a shortcut that can be used when all of the reactants and products are gases. You may consider volumes as moles. That way it looks like this.
11.2 L H2 x (2 moles NH3/3 mole H2) = 11.2 x (2/3) = xx L NH3.
Then convert xx L NH3 to grams remembering that 22.4 L NH3 has a mass of 17 grams (1 mole).