College General Chemistry: any help?

Question

Methanol, CH3OH, can be made by combining gaseous carbon monoxide and hydrogen gas.
In one experiment 2.07 g of H2 is mixed with 16.90 g of CO.


How many grams of CH3OH will be produced?

Answer 1

First balance the equation to find ratios for the reactants:

2 H2 + CO ---> CH3OH

Now determine the limiting reagent, if any.

2.07g/2.02g/mol = 1.02 moles H2
16.9g/28.01g/mol = 0.601 moles CO

Since we need 2 moles of H2 for each mole of CO we can see that hydrogen will run out first, and we will only get 1.02/2 or 0.510 moles of methanol from this reaction.

Now convert to grams of methanol.

0.510mole x [12.01 + 16.0 + (4 x 1.01)]g/mol =

0.510 mole x 32.05g/mol = 16.35g

Answer 2

It's a limiting reagent problem. I don't have a calculator on me right now so I can't give you the exact answer but:

You need to convert each of the two reactants to moles so:

2.07 g / 2.02 g/mol = however many moles H2

and

16.90 g / 28.01 (i think?) g/mol = however many moles CO

whichever is SMALLER is the limiting reagent. Then based on the balanced reaction:

2 H2 + CO --> CH3OH

you can determine how you need to stoichiometricly change that number. If H2 were your limiting reagent, then by the reaction you would need to divide the number in half because you get half as much methanol as you start with hydrogen. If it were carbon monoxide you'd leave it alone because they appear in equal amounts.

Then you just convert to grams using the molecular weight of methanol.

Hope it helped!