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?
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?