In cold snowy areas water is added to the atmosphere through sublimation condensation transpiration precipitation

3.2 g

Step-by-step explanation:  

We are given the mass of two reactants, so this is a limiting reactant problem.

We know that we will need mases, moles, and molar masses, so, lets assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r:  28.01  2.016       32.04  

          CO  +  2H₂ ⟶ CH₃OH

m/g:    2.8     0.50

Step 1.  Calculate the moles of each reactant  

Moles of CO = 2.8 g   × 1mol/28.01 g  = 0.100 mol

Moles of H₂ = 0.50 g × 1 mol/70.91 g = 0.248 mol

Step 2. Identify the limiting reactant  

Calculate the moles of CH₃OH we can obtain from each reactant.

From CO:  

The molar ratio of CH₃OH:CO is 1:1 .

Moles of CH₃OH = 0.100× 1/1  

Moles of CH₃OH = 0.100 mol CH₃OH

From H₂:  

The molar ratio of CH₃OH: H₂ is 1:2.  

Moles of CH₃OH = 0.248 × 1/2  

Moles of CH₃OH = 0.124 mol CH₃OH  

CO is the limiting reactant because it gives the smaller amount of CH₃OH.  

Step 3. Calculate the theoretical yield.

Theor. yield = 0.0100 mol CH₃OH × 32.04 g CH₃OH /1 mol CH₃OH

Theor. yield = 3.2 g CH₃OH