\(m_{bình1}=m_{H_2O}=6.3\left(g\right)\)
\(\Rightarrow n_{H_2O}=\dfrac{6.3}{18}=0.35\left(mol\right)\)
\(Xlà:ankan\)
\(\Rightarrow n_{ankan}=n_{H_2O}-n_{CO_2}=0.1\left(mol\right)\)
\(\Rightarrow n_{CO_2}=0.35-0.1=0.25\left(mol\right)\)
\(n_{CaCO_3}=n_{CO_2}=0.25\left(mol\right)\)
\(m_{CaCO_3}=0.25\cdot100=25\left(g\right)\)
m bình 1 tăng = \(m_{H_2O}=6,3\left(g\right)\)
\(\rightarrow n_{H_2O}=\dfrac{6,3}{18}=0,35\left(mol\right)\)
\(n_{ankan}=n_{H_2O}-n_{CO_2}\)
\(\rightarrow n_{CO_2}=0,35-0,1=0,25\left(mol\right)\)
\(m_{BaCO3}=0,25.197=49,25\left(g\right)\)
