It is known that 1.0 L of nitrogen gas reacts with 3.0 L estequilometricamente gas ...?
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Hydrogen to form 2.0 L of ammonia gas in the same conditions of pressure and temperature.
a)What is the final volume of gas?
b)5.0 L of N2 + 9.0 L H2. What final volume?
To determine the final volume of gas, we first need to understand the stoichiometry of the reaction between nitrogen gas (N2) and hydrogen gas (H2) to form ammonia gas (NH3). According to the balanced chemical equation:
N2 + 3H2 → 2NH3
From the equation, we can see that for every 1 mole of N2, we need 3 moles of H2 to produce 2 moles of NH3. Using this information, we can calculate the final volume.
a) What is the final volume of gas?
Since we are given the initial volume of N2 as 1.0 L, we can use the stoichiometry to determine the volumes of H2 and NH3.
From the balanced equation, for every 1 L of N2, 3 L of H2 are required. So, if we started with 1.0 L of N2, we would need 3.0 L of H2.
Next, we need to determine the volume of NH3 produced. From the equation, 2 moles of NH3 are produced for every 1 mole of N2. Since the volume ratios are the same as the mole ratios, we can convert the volume of N2 to moles and then convert to the volume of NH3.
Using the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature, we can calculate the number of moles of N2:
n(N2) = PV/(RT) = (1.0 L)(1 atm)/(0.0821 L·atm/mol·K)(T)
Similarly, we can calculate the number of moles of NH3:
n(NH3) = (2 moles N2/1 mole NH3) * n(N2)
To find the final volume of NH3, we can use the ideal gas law again:
V(NH3) = n(NH3)RT/P
Note: The pressure and temperature must remain constant throughout the reaction for these calculations to be valid.
b) 5.0 L of N2 + 9.0 L H2. What final volume?
Using the same approach, we can determine the final volume of gas produced when 5.0 L of N2 and 9.0 L of H2 react. Based on the stoichiometry, we can determine the volume of NH3 produced.
First, we calculate the moles of N2 and H2:
n(N2) = PV/(RT) = (5.0 L)(1 atm)/(0.0821 L·atm/mol·K)(T)
n(H2) = PV/(RT) = (9.0 L)(1 atm)/(0.0821 L·atm/mol·K)(T)
Based on the balanced equation, 2 moles of NH3 are produced for every 1 mole of N2. So, we can determine the number of moles of NH3:
n(NH3) = (2 moles N2/1 mole NH3) * n(N2)
Finally, using the ideal gas law, we calculate the volume of NH3:
V(NH3) = n(NH3)RT/P
Remember to use the same pressure and temperature conditions throughout the calculations for accuracy.
To find the final volume of gas, we need to determine the limiting reactant, which is the reactant that is completely consumed in the reaction.
a) Let's start by calculating the moles of nitrogen gas (N2) and hydrogen gas (H2) in the given reaction:
1 mole of gas at STP (Standard Temperature and Pressure) occupies 22.4 L.
Given:
- 1.0 L of N2
- 3.0 L of H2
Moles of N2 = Volume (L) / 22.4 L/mol = 1.0 L / 22.4 L/mol = 0.0446 mol N2
Moles of H2 = Volume (L) / 22.4 L/mol = 3.0 L / 22.4 L/mol = 0.134 moles H2
The balanced chemical equation for the reaction is:
N2 + 3H2 → 2NH3
According to the stoichiometry of the reaction, 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
Since the stoichiometric ratio is 1:3, the limiting reactant will be N2, as it has fewer moles than H2. This means that N2 will be completely consumed during the reaction.
Now, we can use the stoichiometry of the reaction to find the volume of ammonia gas (NH3) produced:
1 mole of gas at STP occupies 22.4 L.
Moles of NH3 = 2 moles NH3 (according to the stoichiometry)
Using this information, we can calculate the volume of NH3 produced:
Volume of NH3 = Moles of NH3 × 22.4 L/mol
= 2.0 mol × 22.4 L/mol
= 44.8 L
Therefore, the final volume of gas is 44.8 L.
b) To find the final volume when 5.0 L of N2 and 9.0 L of H2 are reacted, we need to determine the limiting reactant and use the stoichiometry of the reaction.
Given:
- 5.0 L of N2
- 9.0 L of H2
Moles of N2 = Volume (L) / 22.4 L/mol = 5.0 L / 22.4 L/mol = 0.2232 mol N2
Moles of H2 = Volume (L) / 22.4 L/mol = 9.0 L / 22.4 L/mol = 0.4027 moles H2
Using the stoichiometry of the reaction, we can determine the limiting reactant:
N2 + 3H2 → 2NH3
According to the stoichiometry, 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
Since the stoichiometric ratio is 1:3, we compare the moles of N2 and H2 to determine the limiting reactant. In this case, N2 has fewer moles, so it is the limiting reactant.
To find the volume of NH3 produced, we can use the stoichiometry of the reaction:
Moles of NH3 = 2 moles NH3 (according to the stoichometry)
Volume of NH3 = Moles of NH3 × 22.4 L/mol
= 2.0 mol × 22.4 L/mol
= 44.8 L
Therefore, the final volume of gas is 44.8 L.