Please answer this BEST ANSWER TONIGHT!!!!!!!!!

Upon heating, calcium carbonate decomposes to calcium oxide and carbon dioxide.
(a) Determine the theoretical yield of CO2 if 248.0 g CaCO3 is heated.
g

(b) What is the percent yield of CO2 if 95.5 g CO2 is collected?
%

CaCO2 = CaO + CO2

molar massof CaCO3 = 100.0 g / mol
moles of CaCO3 = 248.0 / 100.0 = 2.48 moles

from the equation we see the moles of CO2 also = 2.48 moles

theoretical mass of CO2 = 2.48 x 44 g = 109.19

if we have 95.5 g we have 95.5 / 109.19 = 87.5 %

CaCO3 >>>> CaO + CO2 so 1 mole of carbonate makes 1 mole ofCO2

Molar mass of CaCO3 = 100g/mole, so 248g = 2.48 moles and the yield of CO2 will be 2.48 moles

CO2 = 44g/mole so 2.48 moles = 109.12 grams - this is theoretical yield assuming 100 % conversion of reactant to products

If 95.5g of CO2 were collected, the per cent yield is 95.5/ 109.12 X100 = 87.518%

*** a ***

1 CaCO3 ---> 1 CaO + 1 CO2

so... theoretical yield CO2 is...
248.0g CaCO3 x (1 mol CaCO3 / 100.09g CaCO3) x (1 mole CO2 / 1 mol CaCO3) x (44.01g CO2 / mol CO2) = 109.1g CO2...

note the FOUR sig figs?
248.0 has 4
100.09 has 5
44.01 has 4
the rest are exact #'s
since this is multi/div, the result has the same number of sig figs as the factor with the least. 4 !

*** b ***

% yield = actual recovered mass / theoretical mass x 100% = (95.5g / 109.1g) x 100% = 87.5%

note the 3 sig figs in this result. 95.5 has 3 and limits the answer to 3

******
questions?