A.\(300ml=0,3l\)
\(C_{MH_2SO_4}=\dfrac{n}{V}\Rightarrow n_{H_2SO_4}=C_M.V=1.0,3=0,3\left(mol\right)\)
\(PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
Theo PTHH : \(n_{Al}=\dfrac{2}{3}.n_{H_2SO_4}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\)
\(\Rightarrow m_{Al}=n.M=0,2.27=5,4\left(g\right)\)
B. Theo PTHH : \(n_{H_2}=n_{H_2SO_4}=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2}=n.22,4=0,3.22,4=6,72\left(l\right)\)
V = 300 ml = 0,3 l =>md2 = 0,3 kg = 300 g
=> nAl = 0,3.1 = 0,3 mol
2Al + 3H2SO4 -> Al2(SO4)3 + 3H2
0,3 ->0,15 ->0,45
=>VH2 = 0,45 . 22,4 = 10,08 (l)
=>C% = \(\dfrac{0,15.342}{0,3.27+300-0,45.2}\).100% = 16,69%
Mình làm hơi tắt 
( mik k ghi lời giải nha)
