Question #69194

1 Answer
Dec 23, 2015

Answer:

#"4.8 g"#

Explanation:

The trick here is to realize that since the gas is collected over water, you are actually dealing with a gaseous mixture of carbon monoxide and water vapor at a total pressure of #"743 mmHg"#.

In order to get the partial pressure of carbon monoxide in this mixture, you're going to have to find the vapor pressure of water at #25^@"C"# and subtract it from the total pressure of the mixture - think Dalton's Law of partial pressures.

At #25^@"C"#, water has a vapor pressure of about #"23.7 mmHg"#.

http://www.endmemo.com/chem/vaporpressurewater.php

This means that the partial pressure of the carbon monoxide will be equal to

#P_"mixture" = P_(CO) + P_(H_2O)#

#P_(CO_2) = "743 mmHg" - "23.7 mmHg" = "719.3 mmHg"#

From this point on, your strategy will be to use the ideal gas law to find how many moles of carbon monoxide were produced by this decomposition reaction.

#color(blue)(PV = nRT)#

Plug in your values and solve for #n#, the number of moles of gas - do not forget to convert the pressure from mmHg to atm and the temperature from degrees Celsius to Kelvin!

#PV = nRT implies n = (PV)/(RT)#

#n = (719.3/760color(red)(cancel(color(black)("atm"))) * 2.68color(red)(cancel(color(black)("L"))))/(0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K")))) = "0.1036 moles CO"#

Now, formic acid, #"HCO"_2"H"# undergoes decomposition in the presence of acids (for example, sulfuric acid) to form carbon monoxide, #"CO"#, and water, #"H"_2"O"#, in #1:1# mole ratios.

#"HCO"_2"H"_text((aq]) stackrel(color(white)(x)color(red)("H"_2"SO"_4)color(white)(xxx))(->) "H"_2"O"_text((l]) + "CO"_text((g]) uarr#

This means that every mole of formic acid that undergoes decomposition will produce #1# mole of carbon monoxide. So, if the reaction produced #0.1036# moles of #"CO"#, it must have consumed #0.1036# moles #"HCO"_2"H"#.

Now simply use formic acid's molar mass to determine how many grams would contain that many moles

#0.1036 color(red)(cancel(color(black)("moles CHO"_2"H"))) * "46.03 g"/(1color(red)(cancel(color(black)("mole CHO"_2"H")))) = "4.769 g"#

Rounded to two sig figs, the number of sig figs you have for the temperature of the gas, the answer will be

#m_"formic acid" = color(green)("4.8 g")#