If the percentage yield is 92.0% how many grams of CH3OH can be made by reaction of 5.6 x 103g with 1.0 x 103 g

Answer is: 5888 grams of CH₃OH.

Chemical reaction: CO + 2H₂ → CH₃OH.

m (CO) = 5,6·10³ g = 5600 g.

m (H₂) = 1,0·10³ g = 1000 g.

n (CO) = 5600 g : 28 g/mol

n (CO) = 200 mol, limited reactant.

n (H₂) = 1000 g : 2 g/mol.

n (H₂) = 500 mol.

from reaction: n (CO) : n (CH₃OH) = 1 : 1.

n (CH₃OH) = n (CO) = 200 mol.

m (CH₃OH) = 200 mol · 32 g/mol · 0,92.

m (CH₃OH) = 5888 g.