A 2.85-g sample containing only MgO and CuO is placed in a 3.00-L container. The container is filled with CO2 to a pressure of 740. torr at 20. degree C. After the reaction has gone to completion, the pressure inside the flask is 390. torr at 20. degree C. What is the mass percent of MgO in the mixture? Assume that only the MgO reacts with CO2
89.8%
To determine the mass percent of MgO in the mixture, we need to determine the moles of MgO and CuO present.
Step 1: Convert the grams of MgO to moles.
We know that the sample contains only MgO and CuO, so the total mass of the sample is the sum of the masses of MgO and CuO.
Total mass of the sample = mass of MgO + mass of CuO
2.85 g = mass of MgO + mass of CuO
Since we are assuming only MgO reacts with CO2, the final pressure of 390 torr is due to CO2 reacting with MgO.
Step 2: Calculate the moles of CO2 using the ideal gas equation.
PV = nRT
Convert the temperatures to Kelvin: 20. degree C = 20 + 273 = 293 K
R = 0.0821 L atm/(mol K) (gas constant)
Using the ideal gas equation, we can calculate the number of moles of CO2:
(390 torr) * (3.00 L) = n * (0.0821 L atm/(mol K)) * (293 K)
n = (390 torr * 3.00 L) / (0.0821 L atm/(mol K) * 293 K)
Step 3: Calculate the moles of MgO using the balanced chemical equation.
Since MgO reacts with CO2 in a 1:1 ratio, the number of moles of MgO is equal to the number of moles of CO2.
Step 4: Calculate the moles of CuO.
The difference between the total moles of the sample (moles of MgO + moles of CuO) and the moles of MgO is equal to the moles of CuO.
Step 5: Calculate the mass percent of MgO.
Mass percent of MgO = (mass of MgO / mass of the sample) * 100
Using these steps, solve for the mass percent of MgO in the mixture.
To find the mass percent of MgO in the mixture, we need to determine the moles of MgO and CuO, and then calculate the mass percent.
Let's start by calculating the moles of CO2:
Using the ideal gas law, PV = nRT, we can calculate the moles of CO2 in the container before the reaction.
P1 = 740. torr (pressure before reaction)
V1 = 3.00 L (volume of the container)
n1 = moles of CO2
R = 0.0821 L•atm/(mol•K) (gas constant)
T = 20. degree C = 273.15 + 20 = 293.15 K (temperature in Kelvin)
Rearranging the ideal gas law equation, we get:
n1 = P1 • V1 / (R • T)
Substituting the values we have:
n1 = 740. torr • 3.00 L / (0.0821 L•atm/(mol•K) • 293.15 K)
Now we can calculate n1.
Next, let's calculate the moles of CO2 after the reaction:
Using the same formula as above:
P2 = 390. torr (pressure after reaction)
V2 = 3.00 L (volume of the container)
n2 = moles of CO2 after the reaction
n2 = P2 • V2 / (R • T)
Substituting the values we have:
n2 = 390. torr • 3.00 L / (0.0821 L•atm/(mol•K) • 293.15 K)
Now we can calculate n2.
Since only MgO reacts with CO2, we can say that the moles of MgO reacted is equal to the moles of CO2 reacted. This is because the balanced chemical equation for the reaction between MgO and CO2 is 1:1.
Now we find the difference in moles before and after the reaction:
moles of MgO reacted = n1 - n2
Since we know the mass of the sample is 2.85 g and we want to find the mass percent of MgO, we need to know the molar mass of MgO:
Molar mass of MgO = atomic mass of Mg + atomic mass of O
Molar mass of Mg = 24.31 g/mol
Molar mass of O = 16.00 g/mol
Now we can calculate the moles of MgO:
moles of MgO = moles of MgO reacted
Finally, we can calculate the mass percent of MgO in the mixture:
mass percent of MgO = (moles of MgO * molar mass of MgO) / mass of the sample * 100
Substituting the values we have, we can calculate the mass percent of MgO in the mixture.
Let me give you some hints and you take it from there.
Use PV = nRT and calculate moles CO2 in the container initially.
Use PV = nRT and calculate moles CO2 in the container after the reaction is complete.
The difference between before and after should be the moles CO2 that reacted to give MgCO3 with the MgO. That should be equal to moles MgO that reacted (1:1 ratio if you write the equation), change that to grams, and calculate percent from that.