Answer to Question #300510 in Inorganic Chemistry for Nhormelin Gil-o

Question

. How many moles of O2 were needed for the combustion of the alcohol?

b. How many moles of H2O are formed for each mole of O2 consumed?

c. How many grammes of CO2 are produced for each mole of alcohol burned?

d. How many grammes of CO2 are produced for each gram of alcohol burned?

3. How many grams of ethanol, C6H12O is produced from 2000 grams of Glucose, C6H12O6?

Answer 1

Assuming that alcohol is ethanol, the reaction is following:

C₂H₅OH + 3O₂ --> 2CO₂ + 3H₂O

a) 3 moles of O₂ are needed for combustion of 1 mole of alcohol. More details cannot be determined since the amount of alcohol is not specified.

b) 1 mole

c) $m(CO_2)=1\ mol(C_2H_5OH)\times\frac{2\ mol(CO_2)}{1\ mol(C_2H_5OH)}\times\frac{44.01\ g(C_2)}{1\ mol(CO_2)}=88\ g$

Answer: 88 g

d) $m(CO_2)=1\ g(C_2H_5OH)\times\frac{1\ mol(C_2H_5OH)}{46.07\ g(C_2H_5OH)}\times\frac{2\ mol(CO_2)}{1\ mol(C_2H_5OH)}\times\frac{44.01\ g(C_2)}{1\ mol(CO_2)}=1.9\ g$

Answer: 1.9 g

3.

The reaction is below:

C₆H₁₂O₆ --> 2C₂H₅OH + 2CO₂

$m(C_2H_5OH)=2000\ g(C_6H_{12}O_6)\times\frac{1\ mol(C_6H_{12}O_6)}{180.2\ g(C_6H_{12}O_6)}\times\frac{2\ mol(C_2H_5OH)}{1\ mol(C_6H_{12}O_6)}\times\frac{46.07\ g(C_2H_5OH)}{1\ mol(C_2H_5OH)}=1023\ g$

Answer: 1023 g