[johnnyjohn993123]

A chemist weighed 5.14 g of a mixture containing unknown amounts of BaO (s) and CaO(s) (continued below)?
and placed the sample in a 1.5 L flask containing CO₂ (g) at 30°C and 0.987 atm. After the reaction to form BaCO₃ and CaCO₃ was completed, the pressure of CO₂ remaining was 0.303 atm Calculate the mass percents of CaO (s) and BaO(s) in the mixture.

My attemp:
mol of CO₂ reacted with the mixture = n_initial- n_final

n_i=PV/RT
n_i= 0.987 atm * 1.5 L / 0.08206 * 303 K
n_i= 5.95x10^-3 mol CO₂

n_f= 0.303 atm* 1.5 L / 0.08206 * 303 K
n_f=0.0183 mol CO₂

n_reacted= 0.0123 mol CO₂


0.0123 mol CO2 x (1mole BaCO3/1mol CO2) x(197.34g BaCO3/1mole BaCO3) =                              2.43g BaCO3  i.e 47% by mass

0.0123 mol CO2 x (1mole CaCO3/1mole CO2) x (100.087g CaCO3/1moleCaCO3) =            1.23g CaCO3  i.e 24% by mass

I'm Really confused; I dont know how many CO₂ would react on each compound, and I know that it doesn't sum up to a 100% please do help.

[Dan]

My attemp:
mol of CO₂ reacted with the mixture = n_initial- n_final

n_i=PV/RT
n_i= 0.987 atm * 1.5 L / 0.08206 * 303 K
n_i= 5.95x10^-3 mol CO₂

n_f= 0.303 atm* 1.5 L / 0.08206 * 303 K
n_f=0.0183 mol CO₂

n_reacted= 0.0123 mol CO₂

I think the highlighted numbers are incorrect. Note that your number for n_i is smaller than n_f, which does not make logical sense.

[johnnyjohn993123]

well okay , Correction

n_i= 0.0595 mol CO₂
n_reacted = n_i-n_f
n_reacted =  0.0595 - 0.01828
n_reacted = 0.0412 mol CO₂

hence,
There would be:
 8.13 g of BaCO₃ which is 158%
 4.13 g of CaCO₃  which is 80.3 %

I know its wrong but please help me, I've done my attempt already.

[mjc123]

You are asked for the percentages of BaO and CaO, not the carbonates.
You seem to be assuming that all the reacted CO₂ has reacted with both the BaO and the CaO. This is impossible. Some has reacted with BaO, some with CaO.
Suppose you have x g CaO and (5.14 - x)g BaO. How many moles of CO₂ will this mixture react with?