Question

it's a lab work! aim is to determine the mass of CO2 given off in the reaction NaHCO3+HCl gives NaCl+H2O+CO2. i have 2.046 g of NaHCO3 and from that i got the theorotical mass of CO2 (1.07).Mss of HCL is 50.014. i need help in finding the experimental yield like the mass of give out CO2, % yield and error %. Is there any way to get the answer without doing the experiment. i would loose 50 points if i don't turn this in by tomorrow. help me please. Any help would be greatly appreciated

Answer 1

Without the experiment, the moles of CO2 produced can be calculated from the following reaction

The reaction is NaHCO3+ HCl---> NaCl+ H2O+CO2

mass of NaHCO3= 2.046 gm moles of NaHCO3= Mass/molecular weight (84)= 2.046/84=0.024357

Moes of HCl in 50.014 gms =50.014/36.5 ( Molecular weight of HCl)= 1.373

HCl is the excess reactant ( since required is only 0.024357)

Moles of CO2 formed (1mole of NaHCO3 produces 1 mole of CO2 =0.024357

Mass of CO2 produced= 0.024357*44 ( Molecular weight of CO2)= 1.07 gms

Yiled= Actutal mass of CO2/ Theoretical mass of CO2