2H2+3N2-->2NH3
if 28g of H2 gas and 35g of N2 gas are combined, how much NH3 will be produced?
The equation isn't balanced. http://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm
To find the amount of NH3 produced, we need to determine the limiting reactant first.
1. Convert the given masses of H2 and N2 into moles using their respective molar masses:
Mass of H2 = 28g
Molar mass of H2 = 2g/mol
Moles of H2 = 28g / 2g/mol = 14 mol
Mass of N2 = 35g
Molar mass of N2 = 28g/mol
Moles of N2 = 35g / 28g/mol = 1.25 mol
2. Determine the stoichiometric ratio between H2 and NH3 by examining the balanced chemical equation:
2H2 + 3N2 -> 2NH3
From the equation, we can see that for every 2 moles of H2, 2 moles of NH3 are produced.
3. Calculate the theoretical amount of NH3 that can be produced based on the limiting reactant:
The limiting reactant is the one with fewer moles. In this case, N2 has 1.25 moles, while H2 has 14 moles. However, the stoichiometric ratio indicates that it takes three times more moles of N2 to produce the same amount of NH3 compared to H2 (3 moles of N2 for every 2 moles of NH3).
So, we need to calculate how many moles of NH3 can be produced from the available moles of H2:
Moles of NH3 = (14 mol H2) x (2 mol NH3 / 2 mol H2) = 14 mol
Therefore, 14 moles of NH3 can be produced.
4. Convert moles of NH3 into grams:
Molar mass of NH3 = 17g/mol
Mass of NH3 = (14 mol) x (17g/mol) = 238g
Therefore, if 28g of H2 gas and 35g of N2 gas are combined, the maximum amount of NH3 that can be produced is 238g.