The combustion of hexane (C6H14), a component of gasoline, is represented by the balanced chemical equation: 2 C6H14(l) + 19 O2(g) " 12 CO2(g) + 14 H2O(g). If 1.0 mole of hexane undergoescombustion, moles of O2 are required.
2C3H6 (g) + 2NH3 (g) + 3O2 (G) -> 2C3H3N (g) + 6H2O (g)
First off.. not a chem board.. but n e way.
This is a limiting reagent problem.
set it up as a DA problem.(Dimension Analysis)
Start with what you want.
you want Grams of acrylonitrile (C3H3N)
so start with that (Using ACL in place of Acrylonitrile.. just for ease of typing)
(g) = (53 g of ACL/1mol ACL) (2 mols ACL/2 mol C3H6)/ (1mol C3H6/42 grams) (15.0 grams)
solve that you wiill get grams of Acrylonitrile created by 15 grams oc C3H6 = 18.9g
Same setup for the two other reactants.
so i did it and for
oxygen I got 11.04 grams
and for Ammonia i got 15.29 grams
So the most you can make is 11.04 grams because if you have ot make any more .. you will have to get more O2 .. but since you have only 10 grams of it .. that is the most u can make in this reaction.
Both the other reactants are in excess.
rate brainliest pls
Answer:
1 B
2A
#D
3B
1D
2B
3C
4A
A,C,A,B
C,A,C,A
C,A,A,B
Explanation:
Easy ive done this brings back memories
Again, I believe the answer would be D. seawater.
M of CH2O= 12+2+16=30 g/mol
So 30g/mol = 3.5 mol
=> 16g/mol*3.5 divide by 30 = 1.86 mol
I think
