\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: 2K + 2H₂O --> 2KOH + H₂

            0,3<-------------0,3<---0,15

=> m_K = 0,3.39 = 11,7 (g)

=> m_KOH(A) = 21,1 - 11,7 = 9,4 (g)

m_{KOH(dd sau pư)} = 0,3.56 + 9,4 = 26,2 (g)

a = 200 + 0,15.2 - 21,1 = 179,2 (g)

\(C\%=\dfrac{26,2}{200}.100\%=13,1\%\) => x = 13,1 

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(n_{H_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,04\left(mol\right)\Rightarrow m_1=m_{Zn}=0,04.65=2,6\left(g\right)\)

\(n_{HCl}=2n_{H_2}=0,08\left(mol\right)\Rightarrow m_{HCl}=0,08.36,5=2,92\left(g\right)\)

\(\Rightarrow m_2=m_{ddHCl}=\dfrac{2,92}{14,6\%}=20\left(g\right)\)

b, Ta có: m _{dd sau pư} = m_Zn + m _{dd HCl} - m_H2 = 22,52 (g)

\(n_{ZnCl_2}=n_{H_2}=0,04\left(mol\right)\)

\(\Rightarrow C\%_{ZnCl_2}=\dfrac{0,04.136}{22,52}.100\%\approx24,16\%\)

bạn xem thử mình giải vầy đúng không .-.

Theo gt ta có: $n_{Al}=0,1(mol)$

a, $2Al+6HCl\rightarrow 2AlCl_3+3H_2$

b, $\Rightarrow n_{H_2}=0,15(mol)\Rightarrow V_{H_2}=3,36(l)$

c, Ta có: $n_{HCl}=0,3(mol)\Rightarrow m_{HCl}=10,95(g)\Rightarrow \%m_{ddHCl}=219(g)$

d, Bảo toàn khối lượng ta có: $m_{dd}=221,4(g)$

$\Rightarrow \%C_{AlCl_3}=6,02\%$

a)

$2K + 2H_2O \to 2KOH + H_2$

b)

n K = 3,9/39 = 0,1(mol)

Theo PTHH :

n H2 = 1/2 n K = 0,05(mol)

=> V H2 = 0,05.22,4 = 1,12(lít)

c)

m dd sau pư = m K + m nước - m H2 = 3,9 + 36,2 - 0,05.2 = 40(gam)

C% KOH = 0,1.56/40 .100% =14%

a)\(n_K=\dfrac{0,39}{39}=0,01mol\)

\(\left\{{}\begin{matrix}X:KOH\\Y:H_2\end{matrix}\right.\)

b)\(2K+2H_2O\rightarrow2KOH+H_2\)

   0,01      0,01         0,01     0,005

\(V_{H_2}=0,005\cdot22,4=0,112l=112ml\)