This is a limiting reagent problem. How do I know that? Because amounts are give for BOTH reactants. So we must decide which is the limiting reagent. I do this by calculating TWO stoichiometry problems. Here is a reference that will work ALL stoichiometry problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html
moles Mg = 1.25/24.3 = 0.051
moles O2 = 25.0g/32 = 0.78
............2Mg + O2 --> 2MgO
initial....0.050..0.78.....0
change....-0.050.-0.025...0.05
equil......0.....0.75.....0.05
explanation:
initial: You had 0.050 moles Mg and 0.78 moles O2 to start.
change. You will use ALL of the Mg and part of the O2. How much O2 is used? That is 0.05 moles Mg x (1 mole O2/2 moles Mg) = 0.05 x 1/2 = 0.025.
I thing the remaining part of the table should be clear.
How did I know to use -0.05 in the change? Because if I used -.78, under Mg I would have had
Mg...........
0.05
-.78
-?? and we know we can't have a negative number remaining; i.e., the 0.78 is too large so I don't use that one but the smaller one.
Let me know if this is not clear.
refer to the reaction: 2Mg+O2->2MgO. complete the ICE table
2Mg+ O2 2MgO
I 1.25g 25.0g 0g
C ? ? ?
E ? ? ?
I am so lost on the table I have refered to my notes but they are not making any sense. I would really appreciate any help!!
2 answers
No idea lol
1 answer