a, PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(n_{Al}=\dfrac{2,16}{27}=0.08mol\)
\(n_{H_2SO_4}=\dfrac{19.6}{98}=0.2mol\)
Ta thấy: \(\dfrac{0,08}{2}< \dfrac{0.2}{3}\rightarrow H_2SO_4dư\)
Từ phương trình: \(n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,08=0,12mol\)
\(\rightarrow V_{H_2}=22,4.0,12=2,688l\)
b, \(m_{dd_{H_2SO_4}}=\dfrac{19,6.100\%}{10\%}=196g\)
\(m_{H_2}=0,12.2=0.24g\)
\(\rightarrow m_{dd_{spu}}=2,16+196-0,24=197,92g\)
Từ phương trình : \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}.0,08=0.04mol\)
\(\rightarrow m_{Al_2\left(SO_4\right)_3}=0,04.342=13,68g\)
\(\rightarrow C\%_{Al_2\left(SO_4\right)_3}=\dfrac{13,68.100\%}{197,92}\approx7\%\)
Từ phương trình : \(n_{H_2SO_4pu}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0.08=0.12mol\)
\(\rightarrow n_{H_2SO_4dư}=0.2-0.12=0.08mol\)
\(\rightarrow m_{H_2SO_4dư}=0,08.98=7.84g\)
\(\rightarrow C\%_{H_2SO_4dư}=\dfrac{7,84.100\%}{197,92}\approx4\%\)
