The first step to solve this question is to convert the mass of H2 to moles, using its molecular mass:
![2.50gH_2\cdot(molH_2)/(2.00gH_2)=1.25molH_2](https://img.qammunity.org/2023/formulas/chemistry/college/jbhmmv39ntpkeflkzc3hlsh1m70nareyei.png)
Now, convert the volume of CO to moles using the ideal gas law:
![PV=nRT](https://img.qammunity.org/2023/formulas/physics/high-school/ns6tfcyfrork1utxo0g9xitf4bxlxstazf.png)
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):
![\begin{gathered} n=(PV)/(RT) \\ n=(1atm\cdot30.0L)/(0.082atmL/molK\cdot273.15K) \\ n=1.34mol \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/prmp9m6a2mzqtzejmrdykhw7uu4kmcrrl3.png)
From the given equation we can find how many moles of CO react with 1.25 moles of H2:
![1.25molH_2\cdot(1molCO)/(2molH_2)=0.625molCO](https://img.qammunity.org/2023/formulas/chemistry/college/i3mfo10ggujiw2yg8mootdmydki97qceud.png)
Substract this amount to the value of n to find the excess remaining:
![1.34molCO-0.625molCO=0.715molCO](https://img.qammunity.org/2023/formulas/chemistry/college/susuleo1myusnmgdmmrc2y2t4lk21y60dm.png)
Use the ideal gas law once again, but this time to find the volume occupied by 0.715moles of CO:
![\begin{gathered} V=(nRT)/(P) \\ V=(0.715mol\cdot0.082atmL/molK\cdot273.15K)/(1atm) \\ V=16.01L \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/x64pdb5arbowo0sx67u0ikxz6ozcnkzq27.png)
It means that 16.01L of CO are remaining.