HD:

ZnO + 2HCl \(\rightarrow\) ZnCl2 + H2O

Áp dụng định luật bảo toàn khối lượng: 80 + m = 250. Suy ra: m = 170 g.

\(n_{Zn}=\dfrac{13}{65}=0,2mol\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,2      0,4            0,2                  ( mol )

\(m_{HCl}=0,4.36,5=14,6g\)

\(m_{ddHCl}=\dfrac{14,6\times100}{14,6}=100g\)

\(m_{ddspứ}=100+13=113g\)

\(m_{ZnCl_2}=0,2.136=27,2g\)

\(C\%_{ZnCl_2}=\dfrac{27,2}{113}.100=24,07\%\)

a) 

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

_____0,2----------------------->0,2

Xét m_Zn -m_H2 = 13-0,2.2=12,6(g)

=> khối lượng dd sau phản ứng tăng 12,6g

b) 

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PTHH: 2Fe + 6H₂SO₄ --> Fe₂(SO₄)₃ + 3SO₂ + 6H₂O

_____0,1--------------------------------->0,15

Xét m_Fe - m_SO2 = 5,6 - 0,15.64 = -4

=> Khối lượng dung dịch sau phản ứng giảm đi 4g

\(n_{H_2} = \dfrac{15,6-14}{2} = 0,8(mol)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Mg + 2HCl \to MgCl_2 + H_2\)

Gọi \(n_{Al} = a \ mol;n_{Mg} = b\ mol\)

Ta có :

\(\left\{{}\begin{matrix}27a+24b=15,6\\1,5a+b=0,8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,4\\b=0,2\end{matrix}\right.\)

Vậy : 

\(\%m_{Al} = \dfrac{0,4.27}{15,6}.100\% = 69,23\%\\ \%m_{Mg} = 100\% - 69,23\% = 30,77\%\)

Bài 2:

\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\\ n_{Fe}=n_{H_2}=0,45\left(mol\right);n_{HCl}=2.0,45=0,9\left(mol\right)\\ a,m_{Fe}=0,45.56=25,2\left(g\right)\\ b,C_{MddHCl}=\dfrac{0,9}{0,15}=6\left(M\right)\)

a)

$Zn + 2HCl \to ZnCl_2 + H_2$
$ZnO + 2HCl \to ZnCl_2 + H_2O$

b)

Theo PTHH : 

$n_{Zn} = n_{H_2} = \dfrac{4,48}{22,4} = 0,2(mol)$
$\Rightarrow m_{Zn} = 0,2.65 = 13(gam)$

$\Rightarrow m_{ZnO} = m_{hh} - m_{Zn} = 8,8 - 13 = -4,2 < 0$(Sai đề)

Ta có: \(n_{ZnO}=\dfrac{16,2}{81}=0,2\left(mol\right)\)

a, PT: \(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

______0,2_____0,4____0,2 (mol)

b, \(m_{HCl}=0,4.36,5=14,6\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{14,6}{10\%}=146\left(g\right)\)

c, \(C\%_{ZnCl_2}=\dfrac{0,2.136}{16,2+146}.100\%\approx16,77\%\)

Bạn tham khảo nhé!

Zn+2HCl->ZnCl2+H2

0,2--------------0,2

n ZnCl2=27,2\136=0,2 mol

=>m Zn=0,2.65=13g