At STP, 16L of N2 and 48L of H2 are mixed. Assume all the reactants are consumed, how many L of NH3 could be produced? N2(g) +3H2(g) --> 2NH3(g)
To determine how many liters of NH3 could be produced, we need to use the stoichiometry of the balanced equation.
According to the balanced equation: N2(g) + 3H2(g) -> 2NH3(g)
For every 1 mole of N2, 2 moles of NH3 are produced. Therefore, the ratio of N2 to NH3 is 1:2.
We'll start by converting the given volumes of N2 and H2 to moles using the ideal gas law:
PV = nRT
For N2:
n = PV/RT = (1 atm)(16 L)/(0.0821 L*atm/mol*K)(273 K) ≈ 0.6 moles
For H2:
n = PV/RT = (1 atm)(48 L)/(0.0821 L*atm/mol*K)(273 K) ≈ 2.4 moles
From the balance equation, we can see that the amount of NH3 produced is directly proportional to the amount of N2(g) consumed. Therefore, using the mole ratio, we can determine the amount of NH3 produced:
NH3 produced = 2 * (0.6 moles) = 1.2 moles
Finally, we can convert moles of NH3 back to liters at STP:
PV = nRT
V = (nRT)/P = (1.2 moles)(0.0821 L*atm/mol*K)(273 K)/(1 atm) ≈ 27.9 L
Therefore, approximately 27.9 liters of NH3 could be produced.
To solve this question, we need to use the concept of stoichiometry. We can start by finding the limiting reactant:
1. Determine the moles of N2:
- Given volume of N2 = 16 L
- At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters
- Moles of N2 = volume of N2 / volume of 1 mole of N2
= 16 L / 22.4 L/mol
= 0.71 moles of N2
2. Determine the moles of H2:
- Given volume of H2 = 48 L
- Moles of H2 = volume of H2 / volume of 1 mole of H2
= 48 L / 22.4 L/mol
= 2.14 moles of H2
3. Using the balanced chemical equation, determine the stoichiometric ratio between N2 and NH3:
- From the balanced equation: 1 mole of N2 produces 2 moles of NH3
4. Determine the limiting reactant:
- To find the limiting reactant, we compare the mole ratios from steps 1 and 3.
- The mole ratio of N2 to NH3 is 1:2
- The mole ratio of H2 to NH3 is 3:2
- Since the mole ratio of H2 to NH3 is lower than the mole ratio of N2 to NH3, H2 is the limiting reactant.
5. Determine the moles of NH3 that can be produced:
- The stoichiometric ratio between H2 and NH3 is 3:2
- Moles of NH3 = Moles of H2 * (2 moles of NH3 / 3 moles of H2)
= 2.14 moles * (2/3)
= 1.43 moles of NH3
6. Convert moles to volume:
- Since we are given that the reactants are consumed completely, the number of moles of NH3 produced is equal to the number of moles of NH3 that can be collected.
- Given volume = Moles of NH3 * volume of 1 mole of NH3
= 1.43 moles * 22.4 L/mol
= 32.03 L
Therefore, approximately 32.03 liters of NH3 can be produced in this reaction.
To find out how many liters of NH3 can be produced, we need to determine the limiting reactant in the given mixture of N2 and H2. The limiting reactant is the one that is completely consumed first and determines the maximum amount of product that can be formed.
First, we need to convert the given volumes of N2 and H2 to moles using the ideal gas law equation:
PV = nRT
Since the reaction takes place at STP (Standard Temperature and Pressure), we can use the following values for the variables:
- T = 273.15 K
- P = 1 atm
- R = 0.0821 L·atm/(mol·K)
Starting with N2:
Using the given volume of 16L and the ideal gas law equation, we can calculate the number of moles of N2 as follows:
n(N2) = (P * V) / (R * T)
= (1 atm * 16L) / (0.0821 L·atm/(mol·K) * 273.15 K)
≈ 0.608 moles
Similarly, for H2:
Using the given volume of 48L and the ideal gas law equation, we can calculate the number of moles of H2 as follows:
n(H2) = (P * V) / (R * T)
= (1 atm * 48L) / (0.0821 L·atm/(mol·K) * 273.15 K)
≈ 1.776 moles
Now, we need to determine the stoichiometric ratio between N2 and NH3, which is 1:2. It means that for every mole of N2, we can produce 2 moles of NH3.
Since the stoichiometric ratio is 1:3 between H2 and NH3, we need to consider the H2 in excess to find out the limiting reactant.
To do that, we compare the number of moles of N2 and H2, assuming complete consumption of both reactants.
Since we have 0.608 moles of N2, we can produce a maximum of 2 * 0.608 = 1.216 moles of NH3.
Since we have 1.776 moles of H2, we can produce a maximum of 3 * 1.776 = 5.328 moles of NH3.
Comparing the amounts of NH3 that could be produced from the moles of N2 and H2, we can conclude that the limiting reactant is N2, as it generates the smaller quantity of NH3.
Now, we need to convert the moles of NH3 to liters using the ideal gas law equation:
V = (n * R * T) / P
Using the moles of NH3 produced from N2 (1.216 moles) and the ideal gas law equation with the same values for the variables (P, T, and R), we can calculate the volume (V) as follows:
V = (1.216 moles * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm
≈ 27.8 liters
Therefore, approximately 27.8 liters of NH3 can be produced when 16 liters of N2 and 48 liters of H2 are mixed at STP and reacted completely.