\(PTHH:2NaHCO_3+H_2SO_4\rightarrow Na_2SO_4+2CO_2+2H_2O\)
\(n_{CO2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PTHH:
\(n_{NaHCO3}=n_{CO2}=0,1\left(mol\right)\)
\(\rightarrow m_{NaHCO_3}=0,1.84=8,4\left(g\right)\)
Vậy hiệu suất phản ứng là:
\(H=\frac{8,4.100\%}{25,2}=\frac{100\%}{3}\approx33,3\%\)