Limiting Reactants Question

You allow 75g of carbon monoxide to reactive with 58g of hydrogen according to the following process. Determine the limiting reactants and the amount of methanol produced.

CO (g) + H2 (g) -> CH3OH (l)

If possible, I would definitely like to have steps shown. Chemistry isn't my best subject and having the steps would help a lot. Thanks in advance for the help!

Now, my calculations may be wrong, since I took chemistry last year, though I am a tutor so I brushed up on the facts.
NOTE: Your equation isn't balanced correctly. You need more hydrogen on reactant side; should read CO + (2)H2 = CH3OH
First, we have to find out the limiting reactant. To do this, we convert the grams of the reactants to grams of the product--theoretical masses that show how much the product would weigh if the reactant was used as a whole.
75.0 g of CO * 1 mol CO/28.0 g of CO * 1 mol CH3OH/ 1 mol CO * 32.0 g CH3OH/1 mol CH3OH
= 85.71 g of CH3OH

58g H2 * 1 mol H2/2.0g H2 * 1 mol CH3OH/2 mol H2 * 32.0 g CH3OH/ 1 mol CH3OH
= 464g CH3OH
We use the lower number of product produced, because the reaction is limited by a reactant. So, the theoretical yield is 85.71 g, this also shows that CO is our limiting reactant.

Now, we find out how much of the other reactant is left after our CO is used completely up. So, we convert the grams used of limiting reactant (75 g of CO) to figure out how much of the other is used.
75.0 g CO * 1 mol CO/ 28.0gCO * 2 mol H2/1 mol CO * 2.0 g H2/1 mol H2 = 10.71 g H2
We know now that 10.71 grams of H2 is used in this reaction, however we started with 58.0 g. A simple subtraction problem will give us our excess amount:
58.0 g - 10.71 g = 47.29 g H2 in excess.

Hope this helps you understand. :)