The oxides of Group 2A metals (symbolized by M here) react with carbon dioxide according to the following reaction:

MO(s) + CO2(g)  MCO3(s)
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.°C. After the reaction has gone to completion, the pressure inside the flask is 390. torr at 20.°C. What is the mass percent of MgO in the mixture? Assume that only the MgO reacts with CO2.

The CO2 pressure drops from 740 torr to 390 torr; the difference is the pressure of CO2 that reacted and that converted to atm ((740-390)/760 = ?atm

Use PV = nRT to solve for mols CO2.
MgO + CO2 ==> MgCO3.
mols CO2 = mols MgO and that x molar mass = grams MgO.
%MgO = (grams MgO/mass sample)*100 = ?

71.6%

81.3%

(740-390)/760=.46Atm

.46atm(3.0K)=n(.08206)(293.15k)
1.38=n(24.056)
n=.054 mols CO2
mols CO2=mols MgO
g/.054mol=40.31mm of MgO
g=2.177g MgO
2.177gMgO/2.85g total samp
=76.4percent

To find the mass percent of MgO in the mixture, we need to determine the amount of MgO and CuO present in the mixture and then calculate the mass percent.

Let's start by finding the number of moles of CO2 using the ideal gas law equation:

PV = nRT

Where:
P = pressure of CO2 (in atm)
V = volume of the container (in L)
n = number of moles of CO2
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)

First, we need to convert the pressure from torr to atm and the temperature from Celsius to Kelvin.

Pressure: 1 atm = 760 torr
Temperature in Kelvin: 20°C + 273.15 = 293.15 K

Converting the given pressures to atm:
Initial pressure (P1) = 740 torr = 740/760 atm
Final pressure (P2) = 390 torr = 390/760 atm

Now, we can calculate the number of moles of CO2 using the ideal gas law:

n = (PV) / (RT)

n1 = (740/760) * 3.00 / (0.0821 * 293.15)
n2 = (390/760) * 3.00 / (0.0821 * 293.15)

Next, let's determine the amount of MgO that reacts with CO2. According to the balanced equation, 1 mole of MgO reacts with 1 mole of CO2:

n(MgO) = n(CO2)

Now, we calculate the mass of MgO that reacted:

mass(MgO) = n(MgO) * molar mass(MgO)

The molar mass of MgO is:

24.31 g/mol (molar mass of Mg) + 16.00 g/mol (molar mass of O) = 40.31 g/mol

Finally, we can calculate the mass percent of MgO in the mixture:

mass percent(MgO) = (mass(MgO) / mass of mixture) * 100

To find the mass of the mixture, we need to determine the masses of the individual oxides.

We are given that the total mass of the mixture is 2.85 g. By assuming that only MgO reacts with CO2, the mass of CuO can be calculated:

mass(CuO) = total mass - mass(MgO)

Now, we can substitute the values into the formula to get the mass percent of MgO in the mixture:

mass percent(MgO) = (mass(MgO) / (mass(MgO) + mass(CuO))) * 100

By following these steps and substituting the values obtained from the given information, you will be able to calculate the mass percent of MgO in the mixture.