Stoichiometry am i doing it right?

Question

i would appreciate if someone could teach me how with instructions

and reason and check what i did

C6H12O6 -> 2 C2H5OH 2 CO2

a) how many moles of CO2 are produced when .3 moles of C6H12O6 react in this fashion?

i used the ratio 1:2:2 and used .3

i got .6 is that right?

b) how many grams of C6H12O6 are needed to form 2 g of C2H5OH?

c) how many grams of CO2 form when 2 g are C2H5OH are produced?

Answer 1

indeed 0.3 moles of glucose will produce 0.6 moles of CO2

2 grams of ethanol is 2/46 grams= 0.0435 moles

so you will need half that in glucose ( 1:2 ratio glucose/ethanol ) = 0.02175 moles = 3.915 grams of glucose

the amount of CO2 in moles is the same as the ethanol so 0.0435 moles of CO2 will form as well 0.0435 X 44grtams/mole = 1.914 grams  

Answer 2

a) Yes, you get twice as many mols as your starting sugar.

When converting mols/weight, it's easier to visualize if you set up a little table of what you know...

compd................MW......mols....g (=MW x mols)

(A) C6H12O6......180........1........180

(B) C2H5OH.........46........2.........92

(C) CO2...............44........2.........88

b) If 180g of A will make 92g of B, 2g requires 2 x 180/92 = 3.91g

c) If 88g of C form when 92g of B do, 2g form when 2 x 88/92 = 1.91g do.

[ Do not know how many significant figures your full problem calls for...go back and check this, then round off as needed...if you REALLY are using only one sig figure, answers are b)=4, and c)=2 ]

A systematic approach makes this type of problem easier to handle...too bad that tables are difficult to align on this site :-(

Answer 3

C6H12O6 180g/mol

C2H5OH  46g/mol

CO2 44g/mol

from eqn: 1mol*180g/mol=2mol*46g/mol+2mol*44g/mol

180g....92g....................88g

x........2g   ....................y

x=3.913g C6H12O6

y=1.913g C2H5OH

Answer 4

The only thing I would add to the answers you have is that the fundemental principle of Stoichiometry is based on the Law of Conservation of Mass.  Which means that mass or moles of each element is the same on either the left or right of the reaction.