Answer
$32.94 L$ of $H_{2}$ gas is required.
$16.46955475 L$ of CO gas is required.
Work Step by Step
Using the ideal gas law $PV=nRT$ where "P" is the pressure the gas exerts, "V" is the volume the gas occupies, "n" represents the number of moles of the gas, "R" is the ideal gas constant and "T" represents the temperature, we can solve many gaseous questions that behave in ideal conditions. The question asks for the volume of the gas so we must isolate for the volume first from the ideal gas equation to obtain $V = \frac{nRT}{P}$.
First we must convert the pressure in units of atm;
$748 mmHg\times\frac{1 atm}{760 mmHg} = 0.984211 atm$.
We must also convert the temperature into Kelvin by simply adding $273.15$ to the Celcius value; $86^{\circ}C + 273.15 = 359.15 K$
$V = \frac{(0.55 mols)(0.08206 \frac{L\times atm}{mol\times K})(359.15 K)}{(0.984211 atm)} = 16.46955475 L CH_{3}OH $.
Based on the balanced equation; $CO + 2H_{2} -->CH_{3}OH $, we can solve for the volume of $CO$ and $H_{2}$ required by using the stoichiometric coefficients in front of the equations for the respective gases.
$16.46955475 L CH_{3}OH\times\frac{2}{1} = 32.94 L H_{2}$
$16.46955475 L CH_{3}OH\times\frac{2}{1} = 16.46955475 L CO$
