when 25g of ch5oh reacts with 30g of 02

Question

Need the formulas for both

Answer 1

First, we must recognize that this is a combustion reaction and from there predict the products and balance the equation:

 

CH₅OH + 2O₂ --> CO₂ + 3H₂O

 

Given 25 grams of CH5OH, we determine how many moles of CH5OH we have by dividing by its molar mass:
25 g x (1 mole/34 g) = 0.735 moles CH5OH

 

Given 30 grams of O2, we determine how many moles of O2 we have by dividing by its molar mass:
30 g x (1 mole/32 g) = 0.9375 moles O2

 

Given 0.735 moles of CH5OH, we determine how many moles of O2 we would NEED to completely react by multiplying by their mole ratio:

0.735 moles CH5OH x (2 moles O2/1 moles CH5OH) = 1.47 moles O2

 

Needing 1.47 moles of O2, we compare that to the moles of O2 that we actually HAVE (0.9375 moles O2) and determine the limiting reactant:

Since would need 1.47 moles of O2 to completely react, but only have 0.9375 moles of O2 present, we will run out of O2 before we run out of CH5OH, making O2 the limiting reactant. It limits how much product can be formed.