NH3(g) + O2(g) NO2(g) + H2O(g)
Consider the above unbalanced equation. What volume of NH3 at 950. mm Hg and 33.5°C is needed to react with 190. mL of O2 at 485 mm Hg and 143°C?
=? mL
To find the needed volume of NH3 gas, we will use the ideal gas law equation PV = nRT. First, we need to gather the necessary information:
1. Convert the given temperatures from Celsius to Kelvin.
33.5°C + 273.15 = 306.65 K (temperature of NH3)
143°C + 273.15 = 416.15 K (temperature of O2)
2. Convert the given pressures to atm.
950. mm Hg * (1 atm / 760 mm Hg) = 1.25 atm (pressure of NH3)
485 mm Hg * (1 atm / 760 mm Hg) = 0.64 atm (pressure of O2)
3. Convert the volumes to liters.
190 mL * (1 L / 1000 mL) = 0.19 L (volume of O2)
Now, we can set up the equation using the ideal gas law:
(P1 * V1) / (n1 * T1) = (P2 * V2) / (n2 * T2)
Where:
P1 = pressure of NH3
V1 = volume of NH3 (what we're trying to find)
n1 = number of moles of NH3
T1 = temperature of NH3 in Kelvin
P2 = pressure of O2
V2 = volume of O2
n2 = number of moles of O2
T2 = temperature of O2 in Kelvin
Since the equation is unbalanced, we need to determine the mole ratio of NH3 to O2 first. From the equation, we can see that the ratio is 1:1, so n1 = n2.
The ideal gas law equation becomes:
(P1 * V1) / (n * T1) = (P2 * V2) / (n * T2)
Since n1 = n2, we can cancel out the n terms.
(P1 * V1) / T1 = (P2 * V2) / T2
Now, substitute the known values into the equation:
(1.25 atm * V1) / 306.65 K = (0.64 atm * 0.19 L) / 416.15 K
Now, rearrange the equation to solve for V1:
V1 = (0.64 atm * 0.19 L * 306.65 K) / (1.25 atm * 416.15 K)
Calculating this expression will give us the volume of NH3 needed, expressed in liters.
1. Balance the equation.
2. Convert 190 mL of O2 at the conditions listed to moles. Use PV = nRT
3. Using the coefficients in the balanced equation, convert moles O2 to moles NH3.
4. Now convert moles NH3 to liters at the conditions listed using PV = nRT, then change to mL.