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.

Adozen is 1221/12 = 1.751.75 x 12 = 21therefore, 1.75 is the answer.

Both models are effective because they show the valence electrons, and these valence electrons indicate the reactivity of an element.