Assuming I have the following reaction:
LiO2 + C +2 Mg -> LiC + 2MgO
I'll be using 0.1 g of LiO2 to react with 1.2 mol C and 2.2 mol Mg in excess. How much C and Mg in grams do I need?
To determine how much C and Mg in grams you need for the reaction, you need to apply stoichiometry, which is the quantitative relationship between the reactants and products in a chemical reaction.
First, let's calculate the molar mass of LiO2:
- LiO2 has a molar mass of 29.88 g/mol (6.94 g/mol for Li and 15.999 g/mol for O2)
Next, let's calculate the number of moles of LiO2 you are using:
- Given that you have 0.1 g of LiO2, divide the mass by the molar mass to get the number of moles:
0.1 g LiO2 / 29.88 g/mol LiO2 = 0.00335 mol LiO2
Based on the balanced equation, we can see that the stoichiometric ratio between LiO2 and C is 1:1. So, you need 0.00335 mol of C.
To calculate the mass of C needed, you need to know its molar mass:
- Carbon (C) has a molar mass of 12.01 g/mol
Multiply the number of moles of C by its molar mass to get the mass:
0.00335 mol C x 12.01 g/mol C = 0.040235 g C
Now, let's move on to magnesium (Mg). From the balanced equation, we can see that the stoichiometric ratio between LiO2 and Mg is also 1:2. However, you mentioned that you have an excess of 2.2 mol Mg, so you don't need to calculate the amount of Mg required.
Therefore, you only need 0.040235 g of C.