The first step in the process used to recover zinc metal from zinc sulfide ore is the reaction of zinc sulfide with oxygen gas to produce zinc oxide and sulfur dioxide.
2 ZnS(s) + 3 O2(g) → 2 ZnO(s) + 2 SO2(g)
When the external pressure is 142700 Pa and the temperature is 689 K, the amount of work performed is 879.1 J. Calculate how many grams of oxygen are consumed in the reaction. i'm not sure how to solve it. Thanks
Dr Bob can you please help me? I'm stuck on how to convert to grams from moles. I used w=PV rearranged it to w/P=V and changed the volume from mL to L. Then I used ideal gas law PV=nRT to find n=moles. Also do i need to convert external pressure Pa to atm when i use PV=nRT to calculate for moles?
To solve this problem, we need to use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure (in Pascals)
V = volume (in liters)
n = number of moles
R = ideal gas constant (8.314 J/(mol·K))
T = temperature (in Kelvin)
First, we need to calculate the volume (V) using the Ideal Gas Law equation. However, we need to know the number of moles (n) of oxygen consumed in the reaction to calculate the volume.
To find the number of moles, we can use the stoichiometry of the balanced chemical equation. According to the balanced equation:
2 ZnS(s) + 3 O2(g) → 2 ZnO(s) + 2 SO2(g)
The stoichiometry tells us that two moles of zinc sulfide react with three moles of oxygen gas. Therefore, the molar ratio of the reaction is:
3 moles O2 : 2 moles ZnS
From the equation, 2 moles of zinc sulfide require 3 moles of oxygen gas for the reaction. Therefore, we can calculate the moles of oxygen gas needed:
(3 moles O2 / 2 moles ZnS) x (mass ZnS / molar mass ZnS)
Next, we can calculate the volume (V) of the oxygen gas using the equation:
V = (nRT) / P
Finally, we can solve for the mass of oxygen consumed (in grams) using the molar mass of oxygen (32.00 g/mol):
mass O2 = moles O2 x molar mass O2
Let's plug in the values and calculate:
Given:
Pressure (P) = 142700 Pa
Temperature (T) = 689 K
Work performed (W) = 879.1 J
First, let's calculate the moles of oxygen gas consumed:
(3 moles O2 / 2 moles ZnS) x (mass ZnS / molar mass ZnS)
Next, let's calculate the volume of the oxygen gas:
V = (nRT) / P
Then, let's calculate the mass of oxygen consumed:
mass O2 = moles O2 x molar mass O2
To calculate the amount of oxygen consumed in the reaction, we can use the ideal gas law equation and the given information about pressure and temperature.
1. Convert the external pressure to units of atm (atmospheres) since the ideal gas law requires pressure in atm.
To convert pascal (Pa) to atm, divide the pressure by 101325 (1 atm = 101325 Pa).
So, 142700 Pa ÷ 101325 Pa/atm = 1.407 atm (rounded to three decimal places).
2. Plug the values into the ideal gas law equation:
PV = nRT
Where:
- P is the pressure in atm,
- V is the volume in liters (which we assume to be 1 in this case since it is not given),
- n is the number of moles of gas,
- R is the ideal gas constant (0.0821 L·atm/(K·mol)),
- T is the temperature in Kelvin.
Rearrange the equation to isolate n (the number of moles of oxygen gas):
n = PV / RT
Substituting the given values:
n = (1.407 atm) / (0.0821 L·atm/(K·mol)) × 689 K
n ≈ 112.674 mol (rounded to three decimal places).
3. Now, we need to calculate the molar ratio between oxygen (O2) and zinc sulfide (ZnS) in the balanced equation.
From the balanced equation: 2 ZnS(s) + 3 O2(g) → 2 ZnO(s) + 2 SO2(g)
The mole ratio between O2 and ZnS is 3:2.
To find the moles of oxygen gas consumed in the reaction, we need to convert the moles of zinc sulfide to oxygen using the mole ratio:
Moles of O2 = Moles of ZnS × (3 mol O2 / 2 mol ZnS)
Substituting the value calculated in step 2:
Moles of O2 = 112.674 mol × (3 mol O2 / 2 mol ZnS) ≈ 169.011 mol (rounded to three decimal places).
4. Convert moles of oxygen gas (O2) to grams using the molar mass of oxygen gas (O2).
The molar mass of oxygen gas (O2) is approximately 32 g/mol.
Grams of O2 = Moles of O2 × Molar mass of O2
Grams of O2 = 169.011 mol × 32 g/mol ≈ 5408.352 g (rounded to three decimal places).
Therefore, approximately 5408.352 grams of oxygen gas are consumed in the reaction.