Blog Week 29

On Monday, I was absent. On Tuesday, we worked on reactant equations. We watched Ethanol react with something and oxygen. Then we tested to see if it would combust and it did. A combustion reaction makes carbon dioxide and water.

Combustion -> CO2 + H2O. We had an equation for our experiment it was: C2H6O + 3O2, we found the molar mass for C2H6O = 46.08 g/mol and the molar mass for 96 g/mol. So, 46.08 + 96 = 142.08 g/mol. Then we found the reaction equation C2H6O + 3O2 -> 2CO2 + 3H2O. The molar mass for 2CO2 is 88.02 g/mol and the molar mass for 3H20 is 54.06 g/mol. So, 88.02 + 54.06 = 142.08 g/mol. So, the reaction equation is right because the molar masses are equal from before and after the reaction. C2H6O + 3O2 -> 2CO2 + 3H2O, so for every 1 C2H6O there's three H20. Then for every 50 C2H6O there's 15o H2O. We also, went over the equation for g<--> mol, they are g x 1mol/__g = __mol and mol x __g/1mol =__g. On Wednesday, we learned how to do BCA tables. We did a practice:
               2H2S + 3O2 --> SO2 + 2H2O <-- need 3O2 to get rid of 2H2S.
Before:   +2.4       XS          Ø           Ø     <--- Ø because it's being produced 
Change:  -2.4       -3.6        +2.4      +2.4  <-- are mols  
________________________________  <-- 3.6 because it's a 2:3 ratio, (3/2) x 2.4 = 3.6 
After:        0           XS        +2.4      +2.4

After that we wanted to find out how much (g) of H20 did we produce?
mol x molar mass/1 mol = g
2.4mol H2O x 18g/1 mol = 43.2 g.
    ^                      ^
in equation      H2=2 and O = 16 H2O = 18

On Thursday, we did more examples:

2.            C3H8 + 5O2 --> 3CO2 + 4H2O <-- needed to balance equation

Before:       +4       XS          Ø           Ø       <--- Ø because it's being produced 

Change:      -4       -20        +12        +16     <-- are mols  

________________________________    <-- 4 bc it's a 1:3 ratio, (1/3) x 12 = 4; 20 bc 1:5, (1/5) x 4 = 20

After:           0         XS        +12      +16          16 bc 1:4, (1/4) x 4 = 16

6.           3H2 + N2 --> 2NH3 <-- need 3H2 to get rid of N2.

Before:   .45     .15           Ø     <--- Ø because it's being produced 

Change:  -.45    -.15       +.30  <-- are mols  

________________________<-- .45 bc it's a 3:1, (3/1) x .15 = .45 and .30 bc it's a 2:1, (2/1) x .15 = .30   

After:       0         0          +.30

Then we wanted to find how much (g) of 2NH3 we produced?

molar masses N=14 H3=3 

17g/1mol x .30 = 5.1 g of 3NH3.

On Friday, we took an assessment. I think I did okay, but I didn't know the type of chemical reaction. Then I took the assessment from last week. I understood everything, except the chemical reactions.