Answer:
a. 0.8 moles of CO₂ are produced when 0.400 mol of C₆H₁₂O₆ reacts.
b. 7.5 grams of C₂H₅OH are produced by 14.67 grams of C₆H₁₂O₆.
c. 7.17 grams of CO₂ form when 7.50 g of C₂H₅OH are produced.
Explanation:
The balanced reaction is:
C₆H₁₂O₆ → 2 C₂H₅OH + 2 CO₂
By reaction stoichiometry, the following amounts of moles of each compound participate in the reaction:
The molar mass of each compound is:
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
Then you can apply the following rules of three:
a. If 1 mole of C₆H₁₂O₆ produces 2 moles of CO₂, 0.4 moles of C₆H₁₂O₆ produces how many moles of CO₂?
[tex]moles of CO_{2} =\frac{0.4 moles of C_{6} H_{12} O_{6} *2moles of CO_{2} }{1 mole of C_{6} H_{12} O_{6} }[/tex]
moles of CO₂= 0.8 moles
0.8 moles of CO₂ are produced when 0.400 mol of C₆H₁₂O₆ reacts.
b. If 92 grams of C₂H₅OH are produced by 180 grams of C₆H₁₂O₆, 7.5 grams of C₂H₅OH are produced by how many mass of C₆H₁₂O₆?
[tex]mass of C_{6} H_{12} O_{6}=\frac{7.5 grams of C_{2} H_{5} OH* 180 grams of C_{6} H_{12} O_{6}}{92grams of C_{2} H_{5} OH}[/tex]
mass of C₆H₁₂O₆= 14.67 grams
7.5 grams of C₂H₅OH are produced by 14.67 grams of C₆H₁₂O₆.
c. If 92 grams of C₂H₅OH are produced along with 88 grams of CO₂, 7.5 grams of C₂H₅OH are produced along with how many mass of CO₂?
[tex]mass of CO_{2}=\frac{7.5 grams of C_{2} H_{5} OH* 88 grams of C O_{2}}{92grams of C_{2} H_{5} OH}[/tex]
mass of CO₂= 7.17 grams
7.17 grams of CO₂ form when 7.50 g of C₂H₅OH are produced.
