The first step to solve this question is to convert the mass of H2 to moles, using its molecular mass:
[tex]2.50gH_2\cdot\frac{molH_2}{2.00gH_2}=1.25molH_2[/tex]Now, convert the volume of CO to moles using the ideal gas law:
[tex]PV=nRT[/tex]Solve the equation for n and replace P, V, T and R for its known values (1atm, 30.0L, 0.082atmL/molK and 273.15K):
[tex]\begin{gathered} n=\frac{PV}{RT} \\ n=\frac{1atm\cdot30.0L}{0.082atmL/molK\cdot273.15K} \\ n=1.34mol \end{gathered}[/tex]From the given equation we can find how many moles of CO react with 1.25 moles of H2:
[tex]1.25molH_2\cdot\frac{1molCO}{2molH_2}=0.625molCO[/tex]Substract this amount to the value of n to find the excess remaining:
[tex]1.34molCO-0.625molCO=0.715molCO[/tex]Use the ideal gas law once again, but this time to find the volume occupied by 0.715moles of CO:
[tex]\begin{gathered} V=\frac{nRT}{P} \\ V=\frac{0.715mol\cdot0.082atmL/molK\cdot273.15K}{1atm} \\ V=16.01L \end{gathered}[/tex]It means that 16.01L of CO are remaining.
