When calcium carbonate is heated, it produces calcium oxide and carbon dioxide. The equation for the reaction is CaCO3(s) CaO(s) CO2(g). How many grams of calcium carbonate (molar mass = 100 g/mol) need to decompose to produce 44.5 g of Calcium oxide.
Answer: 79.5 grams
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP, contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles and weighs equal to the molecular mass of the substance.
[tex]CaCO_3(s)\rightarrow CaO(s)+CO_2(g)[/tex]
According to stoichiometry :
1 mole of CaO is produced by 1 mole of [tex]CaCO_3[/tex]
That means 56 g of CaO is produced by 100 g of [tex]CaCO_3[/tex]
Thus 44.5 g of CaO is produced by =[tex]\frac{100}{56}\times 44.5=79.5g [/tex] of [tex]CaCO_3[/tex]
Thus 79.5 g of [tex]CaCO_3[/tex] need to decompose to produce 44.5 g of Calcium oxide.
