2AL + 3H2SO4 => Al2(SO4)3 + 3H2
2 3 1 3
0.05
Đổi: 200ml = 0,2l
nH2=1.12/22.4=0.05 mol
=>nH2SO4=0,05*3/3=0,05 mol
=> CM = n / V = 0.05 / 0,2 = 0,25M
Done!
Không biết đúng không nữa :v
\(V_{dd\left(H_2SO_4\right)}=200\left(ml\right)=0,2\left(l\right)\)
\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
\(2Al+3H_2SO_4\underrightarrow{t^o}Al_2\left(SO_4\right)_3+3H_2\)
mol 2 3 1 3
mol 0,03 0,05 0,02 0,05
\(C_{M\left(H_2SO_4\right)}=\dfrac{n}{V_{dd}}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
\(V_{dd\left(H_2SO_4\right)}=200\left(ml\right)=0,2\left(l\right)\)
