a) \(n_{SO_3}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PTHH: SO3 + H2O --> H2SO4
0,2------------>0,2
\(m_{H_2SO_4}=0,2.98=19,6\left(g\right)\)
\(m_{dd.sau.pư}=0,2.80+144=160\left(g\right)\)
=> \(C\%=\dfrac{19,6}{160}.100\%=12,25\%\)
b) \(V_{dd}=\dfrac{160}{1,25}=128\left(ml\right)=0,128\left(l\right)\)
=> \(C_M=\dfrac{0,2}{0,128}=1,5625M\)
c)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
\(\dfrac{0,4}{3}\)<--0,2
=> \(m_{Al}=\dfrac{0,4}{3}.27=3,6\left(g\right)\)
\(n_{SO_3}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ m_{SO_3}=0,2.80=16\left(g\right)\\ m_{dd}=16+144=160\left(g\right)\)
PTHH: SO3 + H2O ---> H2SO4
0,2---------------->0,2
\(\rightarrow m_{H_2SO_4}=0,2.98=19,6\left(g\right)\\ \rightarrow C\%_{H_2SO_4}=\dfrac{19,6}{160}.100\%=12,25\%\)
Ta có: \(C_M=C\%.\dfrac{10D}{M}\)
\(\rightarrow C_{M\left(H_2SO_4\right)}=12,25.\dfrac{10.1,25}{98}=1,5625M\)
PTHH: 3H2SO4 + 2Al ---> Al2(SO4)3 + 3H2
0,2------>\(\dfrac{2}{15}\)
=> \(m=\dfrac{2}{15}.27=3,6\left(g\right)\)
