I'm stumped
4ko2 + 2co2 ----2k2co3 + 3o2
What mass of k02 in grams is required to react with 8.90 l at 22.0 c and 767 mm hg?
Please help thank you
Please note that you will not get anywhere in chemistry, probably flunk out, it you don't find and use the caps key. You must know where it is for you started sentences with a capital letter and I must congratulate you on ending the sentences properly BUT lower case and upper case letters in chemistry mean different things. For example, CO vs Co vs co OR m vs M. Also note that you didn't react the KO2 with anything. I assume you meant 8.90 L CO2. If not ignore the rest of this.
Use PV = nRT and convert 8.90L CO2 to n = mols CO2.
Using the coefficients in the balanced equation, convert mols CO2 to mols KO2.
Then convert mols KO2 to grams. g KO2 = mols KO2 x molar mass KO2.
To solve this problem, we need to use the ideal gas law equation:
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature (in Kelvin)
First, let's convert the given information into the proper units:
- Volume: 8.90 L
- Temperature: 22.0°C = 22.0 + 273.15 = 295.15 K
- Pressure: 767 mmHg = 767/760 atm (approximately) = 1.0092 atm
Now, let's rearrange the ideal gas law equation to solve for the number of moles (n):
n = PV / RT
Substituting the values we have:
n = (1.0092 atm) * (8.90 L) / (0.0821 L·atm/(mol·K) * 295.15 K)
Calculating this, we get:
n ≈ 0.3988 mol
Next, we use the balanced chemical equation:
4K2O2 + 2CO2 → 2K2CO3 + 3O2
From the equation, we can see that:
4 moles of K2O2 react to produce 2 moles of K2CO3
To find the moles of K2O2 required for the reaction, we use the following proportion:
4 moles K2O2 / 2 moles K2CO3 = x moles K2O2 / 0.3988 mol K2CO3
Simplifying this proportion:
x ≈ (4/2) * 0.3988 mol K2O2
x ≈ 0.7976 mol K2O2
Finally, we can convert the moles of K2O2 to grams by using the molar mass of K2O2, which is approximately 214.23 g/mol:
Mass of K2O2 = 0.7976 mol * 214.23 g/mol
Mass of K2O2 ≈ 170.22 g
Therefore, approximately 170.22 grams of K2O2 is required to react with 8.90 L of gas at 22.0°C and 767 mmHg pressure.
To solve this problem, you need to use the ideal gas law equation to find the number of moles of oxygen gas (O2) produced in the reaction. Then, you can use stoichiometry to determine the number of moles of potassium oxide (K2O2) required to produce that amount of oxygen gas. Finally, you can convert the number of moles to grams using the molar mass of K2O2.
Step 1: Find the number of moles of O2 produced using the ideal gas law equation:
PV = nRT
Where:
P = pressure in atm
V = volume in liters
n = number of moles
R = ideal gas constant (0.0821 L atm / mol K)
T = temperature in Kelvin
Convert the provided measurements to the appropriate units:
Pressure (P) = 767 mmHg * 1 atm / 760 mmHg = 1.010 atm
Volume (V) = 8.90 L
Temperature (T) = 22.0 °C + 273.15 = 295.15 K
Now, plug the values into the equation:
(1.010 atm) * (8.90 L) = n * (0.0821 L atm / mol K) * (295.15 K)
Solve for n:
n = (1.010 atm * 8.90 L) / (0.0821 L atm / mol K * 295.15 K)
n ≈ 0.385 mol of O2
Step 2: Use stoichiometry to determine the number of moles of K2O2 required to produce the calculated amount of O2.
From the balanced chemical equation, the stoichiometric ratio between K2O2 and O2 is 4:3. This means that for every 4 moles of K2O2, you would produce 3 moles of O2.
Let's set up a ratio:
4 mol K2O2 / 3 mol O2 = x mol K2O2 / 0.385 mol O2
Cross multiply and solve for x (the number of moles of K2O2):
x = (4 mol K2O2 * 0.385 mol O2) / 3 mol O2
x ≈ 0.513 mol K2O2
Step 3: Convert the number of moles of K2O2 to grams:
To convert moles to grams, you need to multiply the number of moles by the molar mass of K2O2.
Molar mass of K2O2 = (2 * atomic mass of K) + atomic mass of O + (2 * atomic mass of O)
Molar mass of K2O2 = (2 * 39.10 g/mol) + 16.00 g/mol + (2 * 16.00 g/mol)
Molar mass of K2O2 = 122.20 g/mol
Now, calculate the mass of K2O2:
mass = number of moles * molar mass
mass = 0.513 mol * 122.20 g/mol
mass ≈ 62.75 g
Therefore, approximately 62.75 grams of K2O2 is required to react with 8.90 L at 22.0 °C and 767 mmHg.