Asked by Helga
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.
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.
Answers
Answered by
DrBob222
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 = ?
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 = ?
Answered by
Anonymous
71.6%
Answered by
Anonymous
81.3%
Answered by
Jt
(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
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