a)Ba+H2SO4--->BaSO4+H2
\(m_{H2SO4}=\frac{100.14,7}{100}=14,7\left(g\right)\)
\(n_{H2SO4}=\frac{14,7}{98}=0,15\left(mol\right)\)
\(n_{Ba}=\frac{27,4}{137}=0,2\left(mol\right)\)
=>Ba dư
\(n_{H2}=n_{H2SO4}=0,15\left(mol\right)\)
\(VH2=0,15.22,4=3,36\left(l\right)\)
b)k có dd sau pư..bn xem lại đề nha