\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
a.
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
0,2--------------------------0,1----------0,3
\(V_{H_2}=0,3\cdot22,4=6,72l\)
b. \(CM_{Al_2\left(SO_4\right)_3}=\dfrac{0,1}{0,2}=0,5M\)
Theo bài ta có
nAl=5,4/27=0,2(mol)
pthh 2Al+3H2SO4=>Al2(SO4)3+3H2
Theo pthh và bài ta có
+) nH2=3/2 * nAl=3/2 * 0,2=0,3 (mol)
=>V (H2)=0,3*22.4=6,32(l)
+) nH2SO4=3/2 * nAl=2/3* 0,2=0,3(mol)
=>Cm (dd H2SO4)=0,3/0,2=1,5(M)
Good luck <3
Phần b mình nhầm đề
+) nAl2O3=1/2* nAl=1/2 * 0,2=0,1(mol)
=>Cm(dd Al2(SO4)3 )=0.1/0,2=0,5(M)
a) 2Al + 3H2SO4 \(\rightarrow\) Al2(SO4)3 + 3H2
nAl = \(\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(\Rightarrow n_{H_2}\) = 0,3 (mol)
\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
b) 200 ml = 0,2 lít
\(n_{Al_2\left(SO_4\right)_3}=0,1\left(mol\right)\)
\(\Rightarrow\)CM = \(\dfrac{n}{V}=\dfrac{0,1}{0,2}=0,5M\)
