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 = ?
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.
4 answers
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
.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