PTHH ( I ) : \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\) .

PTHH ( II ) : \(ZnO+2HCl\rightarrow ZnCl_2+H_2O\) .

\(n_{H_2}=\frac{V_{H_2}}{22,4}=\frac{4,48}{22,4}=0,2\left(mol\right)\)

- Theo PTHH ( I ) : \(n_{Zn}=n_{ZnCl_2}=n_{H_2}=0,2\left(mol\right)\)

=> \(m_{Zn}=n.M=0,2.65=13\left(g\right)\)

=> \(m_{ZnCl_2}=n.M=0,2.\left(65+35,5.2\right)=27,2\left(g\right)\)

Mà sau phản ứng chỉ thu được \(ZnCl_2\) .

=> m_{muối khan} \(=m_{ZnCl_2\left(I\right)}+m_{ZnCl_2\left(II\right)}=27,2+m_{ZnCl_2\left(II\right)}=40,8\)

=> \(m_{ZnCl_2\left(II\right)}=13,6\left(g\right)\)

=> \(n_{ZnCl2}=\frac{m}{M}=\frac{13,6}{65+35,5.2}=0,1\left(mol\right)\)

Theo PTHH ( II ) : \(n_{ZnO}=n_{ZnCl_2}=0,1\left(mol\right)\)

=> \(m_{ZnO}=n.M=0,1.\left(16+65\right)=8,1\left(g\right)\)

a) \(n_{Al} = \dfrac{5,4}{27} = 0,2(mol)\)

\(2Al + 6HCl \to 2AlCl_3 + 3H_2\)

Theo PTHH : \(n_{H_2} = \dfrac{3}{2}n_{Al} = 0,3(mol)\)

\(\Rightarrow V_{H_2} = 0,3.22,4 = 6,72(lít)\)

b)

Gọi \(C\%_{HCl}= a\%\)

Theo PTHH :

\(n_{HCl} = 2n_{H_2} = 0,6(mol)\\ \Rightarrow m_{dd\ HCl} = \dfrac{0,6.36,5}{a\%} = \dfrac{2190}{a}(gam)\)

Sau phản ứng,

\(m_{dd} = m_{Al} + m_{dd\ HCl} - m_{H_2}\\ = 5,4 + \dfrac{2190}{a} - 0,3.2= 4,8 + \dfrac{2190}{a}(gam)\)

1. Đặt kim loại hóa trị 2 là X

ta có phương trình hoá học

X+2HCl → XCl2+H2 (áp dụng quy tắc nhân chéo )

9,75_______20,4

X_______X+70

\(\frac{9,75}{X}=\frac{20,4}{X+70}\)

⇒X≈64 ( Cu)

Cu+2HCl →CuCl2 +H2

0,15 _____________ 0,15

\(n_{Cu}=\frac{9,75}{64}=0,15\left(mol\right)\)

\(\rightarrow V_y=0,15.22,4=3,36\left(l\right)\)

\(a,PTHH:K_2O+H_2O\to 2KOH\\ n_{K_2O}=\dfrac{18,8}{94}=0,2(mol)\\ \Rightarrow n_{KOH}=0,4(mol)\\ \Rightarrow C\%_{KOH}=\dfrac{0,4.56}{18,8+121,2}.100\%=16\%\\ b,n_{KOH}=\dfrac{50.16\%}{56}=\dfrac{1}{7}(mol)\\ PTHH:2KOH+H_2SO_4\to K_2SO_4+2H_2O\\ \Rightarrow n_{H_2SO_4}=\dfrac{2}{7}(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{\dfrac{2}{7}.98}{20\%}=140(g)\)

\(n_{K_2SO_4}=n_{KOH}=\dfrac{1}{7}(mol)\\ \Rightarrow m_{K_2SO_4}=\dfrac{1}{7}.174=24,86(g)\\ \Rightarrow C\%_{K_2SO_4}=\dfrac{24,86}{50+140}.100\%=13,08\%\)

số mol dư nha

P = F/S

CT 

P=d.S.h/S

P:pascal(Pa)

d:N/m³

h: mét

a) PTHH: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

b+c) Ta có: \(n_P=\dfrac{22,4}{31}=\dfrac{112}{155}\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{P_2O_5}=\dfrac{56}{155}\left(mol\right)\\n_{O_2}=\dfrac{28}{31}\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{P_2O_5}=\dfrac{56}{155}\cdot142\approx51,3\left(g\right)\\V_{O_2}=\dfrac{28}{31}\cdot22,4\approx20,23\left(l\right)\end{matrix}\right.\)

c) Ta có: \(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{\dfrac{112}{155}}{4}>\dfrac{0,3}{5}\) \(\Rightarrow\) Photpho còn dư, Oxi p/ứ hết