Question

Question #69194

Answer 1

`"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")`