how many grams of calcium carbonate are required to prepare 41.386 g of calcium oxide

Question

ramzy197 · Answer

so first you have to write a balanced equation so you know the stoichiometry, the equation for that reaction would be:
CaCO3= CaO + CO2
as you can see the equation is balanced and that one mole of calcium carbonate forms one mole of calcium oxide. 
you then have to work out the moles of calcium oxide present: MR Ca= 40 O= 16
mole of calcium oxide = mass/ mr 
                                 = 41.386 /(40+16) 
                                 = 0.74
as i said before 1 mole of calcium carbonate produces one mole of calcium oxide so the moles of calcium carbonate in this reaction would be the same as that of calcium oxide. To work out the mass of calcium carbonate used you do:
mass of calcium carbonate= moles x mr 
                                       = 0.74 x (40+12+(3x16))
                                       = 74
so 74 g of calcium carbonate was used to produce 41.386 g of calcium oxide :)i hope it helped 
if theres any questions you can ask me