12.2 cắt quá bạn đăng lại nhé
13
\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
a.
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,4->1,2----->0,4-------->0,6
\(V_{H_2}=0,6.22,4=13,44\left(l\right)\)
b. \(C\%_{HCl}=\dfrac{1,2.36,5.\left(100+20\right)\%}{200}=26,28\%\)
c. \(n_{HCl.dư}=(\dfrac{200.26,28\%}{100\%}:36,5)-1,2=0,24\left(mol\right)\)
\(C\%_{HCl.dư}=\dfrac{0,24.36,5.100\%}{10,8+200-0,6.2}=4,18\%\)
\(C\%_{AlCl_3}=\dfrac{0,4.133,5.100\%}{10,8+200-0,6.2}=25,48\%\)
14
\(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
0,1---->0,1--->0,1-------->0,05
a. \(V=V_{H_2}=0,05.22,4=1,12\left(l\right)\)
b, \(CM_{NaOH}=\dfrac{0,1}{0,1}=1M\)
15
\(n_K=\dfrac{3,9}{39}=0,1\left(mol\right)\)
\(2K+2H_2O\rightarrow2KOH+H_2\)
0,1-->0,1---->0,1------->0,05
a. \(V_{H_2}=0,05.22,4=1,12\left(l\right)\)
b. \(CM_{KOH}=\dfrac{0,1}{0,2}=0,5M\)
16
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\)
0,1--->0,2----->0,1-------->0,1
a. \(m_{Ca}=40.0,1=4\left(g\right)\)
b, \(CM_{Ca\left(OH\right)_2}=\dfrac{0,1}{0,3}=\dfrac{1}{3}M\)
