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!!
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.
No idea lol
To complete the ICE (Initial, Change, Equilibrium) table for the reaction 2Mg + O2 -> 2MgO, we need to track the initial amounts, the changes that occur during the reaction, and the final equilibrium amounts.
Given:
Initial amount of 2Mg = 1.25g
Initial amount of O2 = 25.0g
Initial amount of 2MgO = 0g
Let's start by filling in the initial amounts (I column) of the ICE table:
2Mg+: 1.25g
O2: 25.0g
2MgO: 0g
Next, let's determine the changes (C column) that occur during the reaction. The equation shows that 2 moles of Mg react with 1 mole of O2 to produce 2 moles of MgO. Therefore, we can use the molar masses of the substances to calculate the number of moles involved.
The molar mass of Mg is approximately 24.31g/mol, and since we have 1.25g, we can calculate the number of moles:
1.25g / 24.31g/mol = 0.051 moles
The molar mass of O2 is approximately 32.00g/mol, and since we have 25.0g, we can calculate the number of moles:
25.0g / 32.00g/mol = 0.781 moles
We can assume that both Mg and O2 are completely consumed in the reaction since they react in a 1:1 stoichiometric ratio.
Therefore, the changes are as follows:
2Mg+: -2(0.051) = -0.102 moles
O2: -0.781 moles
2MgO: +2(0.051) = +0.102 moles
Now we can fill in the changes (C column) of the ICE table:
2Mg+: -0.102 moles
O2: -0.781 moles
2MgO: +0.102 moles
Finally, to determine the final equilibrium amounts (E column), we need to add the initial amounts and changes:
2Mg+: 1.25g - 0.102 moles = ? (You need to convert back to grams)
O2: 25.0g - 0.781 moles = ? (You need to convert back to grams)
2MgO: 0g + 0.102 moles = ? (You need to convert back to grams)
To calculate the values in grams, you can use the given molar masses:
Molar mass of MgO is approximately 40.31g/mol.
Now, use the calculated moles to find the final equilibrium amounts in grams:
2Mg+: calculate using molar mass
O2: calculate using molar mass
2MgO: calculate using molar mass
After finding the final equilibrium amounts (E column) in grams, you can complete the ICE table.
No problem! I can help you with completing the ICE table for the given reaction: 2Mg + O2 -> 2MgO.
An ICE table is used to track the changes in the amount of reactants and products in a chemical reaction. "I" stands for initial, "C" stands for change, and "E" stands for equilibrium. Here's how you can complete the ICE table for this reaction:
I 2Mg + O2 -> 2MgO
I 1.25g 25.0g 0g
In the initial row (I), you simply write down the amounts of each species given in the problem. The coefficients of the balanced reaction equation tell us that there are 2 moles of Mg for every mole of O2 and MgO produced.
To calculate the moles in the initial row, you need to divide the given mass by the molar mass of each species. The molar mass of Mg is 24.31 g/mol, and the molar mass of O2 is 32.00 g/mol.
For Mg: Moles = Mass / Molar Mass = 1.25g / 24.31 g/mol ≈ 0.051 mol
For O2: Moles = Mass / Molar Mass = 25.0g / 32.00 g/mol ≈ 0.781 mol
For MgO: Moles = 0 (since there is no initial MgO)
The next step is to determine the changes (C) in moles for each species. From the balanced equation, you can see that for every 2 moles of Mg reacted, 2 moles of MgO are produced, and for every 1 mole of O2 reacted, 2 moles of MgO are produced.
Therefore, the change in the amount of Mg is -2 times the change in MgO, and the change in the amount of O2 is -1 times the change in MgO.
C ? ? ?
Finally, to find the equilibrium (E) amounts, you need to add the initial (I) and the change (C) amounts for each species.
E ? ? ?
Since the ICE table is incomplete, you will need more information about the changes in order to complete it. Once you have this information, you can calculate the equilibrium amounts by adding the initial and change amounts.