Solution.
ν(NaHCO3)=3.75 mol\nu(NaHCO3) = 3.75 \ molν(NaHCO3)=3.75 mol
ν(H2SO4)=2.11 mol\nu(H2SO4) = 2.11 \ molν(H2SO4)=2.11 mol
ν(NaHCO3)>ν(H2SO4)\nu(NaHCO3)>\nu(H2SO4)ν(NaHCO3)>ν(H2SO4)
ν(H2SO4)=ν(CO2)=2.11 mol\nu(H2SO4) = \nu(CO2) = 2.11 \ molν(H2SO4)=ν(CO2)=2.11 mol
p×V=n×R×Tp \times V = n \times R \times Tp×V=n×R×T
V=n×R×TpV = \frac{n \times R \times T}{p}V=pn×R×T
V=0.05234m3=52.34 LV = 0.05234 m^3 = 52.34 \ LV=0.05234m3=52.34 L
Answer:
V = 52.34 L
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