9. The reaction of 50 mL of N2 gas with 150 mL H2 gas to form ammonia via the equation: N 2(g) + 3H2 (g) → 2NH3 (g) will produce a total of mL of ammonia if pressure and temperature are kept constant
Well, let's do a little math here.
According to the balanced equation, for every 1 mole of N2 gas, you need 3 moles of H2 gas to produce 2 moles of NH3. Since we know the volume of N2 gas is 50 mL and the volume of H2 gas is 150 mL, we can convert these volumes to moles using the Ideal Gas Law:
n = PV/RT
Where:
n = number of moles
P = pressure (constant)
V = volume
R = ideal gas constant
T = temperature (constant)
Since the pressure and temperature are kept constant, we can cancel those out. Now let's calculate the moles of N2 gas and H2 gas:
n(N2) = (50 mL)(1 L/1000 mL) = 0.05 L
n(H2) = (150 mL)(1 L/1000 mL) = 0.15 L
Now, using the stoichiometry of the balanced equation, we can determine the number of moles of NH3 produced:
n(NH3) = (2 moles NH3/1 mole N2)(0.05 moles N2) = 0.1 moles NH3
We can then convert the moles of NH3 to volume using the Ideal Gas Law once again:
V = nRT/P
V = (0.1 moles NH3)(0.0821 L·atm/(mol·K))(273.15 K)/(1 atm) = 2.22245 L
Finally, we can convert this volume to milliliters:
V = (2.22245 L)(1000 mL/L) = 2222.45 mL
So, when 50 mL of N2 gas reacts with 150 mL of H2 gas to produce ammonia, the total volume of ammonia, at constant pressure and temperature, will be 2222.45 mL. That's one big bottle of ammonia! Better hold your nose!
To find the total volume of ammonia produced in the reaction, we first need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
To determine the limiting reactant, we compare the number of moles of each reactant to the stoichiometric coefficients in the balanced equation.
The balanced equation: N2 (g) + 3H2 (g) → 2NH3 (g)
Given:
Volume of N2 gas (V1) = 50 mL
Volume of H2 gas (V2) = 150 mL
Since the volumes are given in mL, we can assume that the reaction is taking place at the same pressure and temperature, allowing us to use volumes directly as a measure of moles.
To find the number of moles of each reactant, we can use the ideal gas law, where PV = nRT.
Rearranging the equation to solve for n (number of moles):
n = PV / RT
Since pressure and temperature are constant, we can ignore them in this calculation.
For N2:
n(N2) = V1
For H2:
n(H2) = V2
Comparing the number of moles of each reactant to the stoichiometric coefficients in the balanced equation:
1 mole of N2 reacts with 3 moles of H2 to form 2 moles of NH3.
Since the balanced equation shows that 1 mole of N2 reacts with 3 moles of H2, we can write the following ratio:
n(N2) / 1 = n(H2) / 3
Substituting the values we determined earlier:
V1 / 1 = V2 / 3
Simplifying the equation:
V1 = (1/3) * V2
Plugging in the given values:
50 mL = (1/3) * 150 mL
Simplifying the equation further:
50 mL = 50 mL
This shows that the volumes are equal, meaning that we have an equal number of moles for both N2 and H2.
Since the two reactants have an equal number of moles, neither is limiting. This means that both reactants will be completely consumed in the reaction.
To find the total volume of ammonia produced, we can use the stoichiometric coefficients from the balanced equation.
The balanced equation shows that 1 mole of N2 reacts to form 2 moles of NH3.
Since we have an equal number of moles of N2 and H2, and each mole of N2 forms 2 moles of NH3, we can conclude that the number of moles of NH3 formed is equal to the number of moles of N2.
Therefore, the total volume of ammonia produced is equal to the volume of N2, which is 50 mL.
Therefore, the reaction will produce a total of 50 mL of ammonia if pressure and temperature are kept constant.
To find the total volume of ammonia produced, we first need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed first, limiting the amount of product that can be formed.
To find the limiting reactant, we compare the moles of each reactant to the stoichiometric coefficients in the balanced equation. The balanced equation tells us that one mole of N2 reacts with three moles of H2 to produce two moles of NH3.
First, we need to convert the volumes of the reactants to moles. To do this, we use the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
Since the pressure and temperature are constant, we can simplify the equation to V/n = constant.
For the N2 gas:
Volume of N2 gas = 50 mL = 0.05 L
Using the ideal gas law, we can calculate the moles of N2:
n1 = (V1/n1) * V1 = constant
n1 = (V1/V2) * n2
n1 = (0.05 L / 22.4 L/mol) * n2 (since 22.4 L/mol is the molar volume of a gas at standard temperature and pressure, which is 1 atm and 273 K)
n1 = 0.002232 moles
For the H2 gas:
Volume of H2 gas = 150 mL = 0.15 L
Using the same approach, we calculate the moles of H2:
n2 = (V1/V2) * n1
n2 = (0.15 L / 22.4 L/mol) * 0.002232 mol
n2 = 0.009396 mol
Now, let's compare the moles of N2 and H2 to determine the limiting reactant:
Moles of NH3 produced = (2/1) * (0.002232 mol) = 0.004464 mol
Since the stoichiometry ratio of N2 to NH3 is 1:2, we know that 0.004464 mol of NH3 corresponds to 0.002232 mol of N2. However, we calculated that we have 0.002232 mol of N2, which means that it is the limiting reactant. This means that all the N2 will be consumed completely, and H2 will be in excess.
According to the stoichiometric ratio, 1 mole of N2 reacts to form 2 moles of NH3. Therefore, the total moles of NH3 produced will be 2 times the moles of N2 consumed:
Total moles of NH3 = 2 * n1 = 2 * 0.002232 mol = 0.004464 mol
Finally, we can convert the moles of NH3 to the volume of NH3 using the molar volume of a gas at standard temperature and pressure:
Volume of NH3 = (0.004464 mol) * (22.4 L/mol) = 0.0999744 L
Converting this volume to milliliters:
Volume of NH3 = 0.0999744 L * 1000 mL/L ≈ 99.97 mL
Therefore, the total volume of ammonia produced, when the pressure and temperature are kept constant, is approximately 99.97 mL.
P V = n R T
V = n (R T/V0
so volume is proportional to number of mos if P and T are constant
we need three mols of H2 for every mol of N2 so there is exactly the right amount of each
for every mol of N2 we will get 2 mol of NH3
so for every mL of N2 we will get 2 mL of NH3
so for 50 mL of N2 we will get 100 mL of NH3