When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction caco3(s)→cao(s)+co2(g) what is the mass of calcium carbonate needed to produce 47.0 l of carbon dioxide at stp? express your answer with the appropriate units?

  The mass  of   calcium  carbonate that  is needed to produce 47.0 l  of Co2  at STP  is  209.8 grams

calculation

CaCO₃ (s) →  CaO(s)  +Co₂

calculate  the  moles  CO2  at STP

AT STP  1  mole  of  gas = 22.4 L

               ?  moles         = 47.0 L

by cross   multiplication

=[ (1 moles  x 47.0  l) / 22.4 L]  =2.098   moles

use of mole ratio  to  calculate the  moles  of CaCO₃

CaCo₃:Co₂  is 1:1  therefore the moles  of CaCo₃ is also  = 2.098  moles

mass = moles  x molar mass

The  molar mass  of CaCo₃ = 40 +12 +(16 x3) =100 g/mol

mass  =2.098 moles  x  100 g/mol =209.8 grams

16

explanation:

the marks go by 4s