a) $Zn+ 2HCl \to ZnCl_2 + H_2$
b) $n_{HCl} = \dfrac{250.7,3\%}{36,5} = 0,5(mol)$
$n_{Zn} = n_{H_2} = \dfrac{1}{2}n_{HCl} = 0,25(mol)$
$m = 0,25.65 =16,25(gam) ; V_{H_2} = 0,25.22,4 = 5,6(lít)$
c)

$m_{dd\ sau\ pư} = 16,25 + 250 - 0,25.2 = 265,75(gam)$
$C\%_{ZnCl_2}  = \dfrac{0,25.136}{265,75}.100\% = 12,8\%$

\(a/ 4Zn+2HCl \to ZnCl_2+H_2 \\ n_{HCl}=\frac{250.7,3\%}{36,5}=0,5(mol)\\ b/ \\ n_{Zn}=n_{H_2}=n_{ZnCl_2}=\frac{1}{2}.n_{HCl}=\frac{1}{2}.0,5=0,25(mol)\\ m_{Zn}=0,25.65=16,25(g)\\ V_{H_2}=0,25.22,4=5,6(l)\\ c/ \\ C\%_{ZnCl_2}=\frac{0,25.136}{16,25+250-0,25.2}.100=12,8\% \)

a)

 
b) 

nZn=nH2=12nHCl=0,25(mol)
m=0,25.65=16,25(gam);VH2=0,25.22,4=5,6(lít)
c)

mdd sau pư=16,25+250−0,25.2=265,75(gam)

PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

Ta có: \(n_{HCl}=0,1\cdot3=0,3\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,1\left(mol\right)\\n_{H_2}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,1\cdot27=2,7\left(g\right)\\V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\end{matrix}\right.\)

\(n_{HCl}=0,1\cdot3=0,3mol\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

 0,1       0,3                        0,15

\(m=0,1\cdot27=2,7g\)

\(V=0,15\cdot22,4=3,36l\)

Câu 1:

Gọi : nMg=a(mol); nMgO=b(mol) (a,b>0)

a) PTHH: Mg + 2 HCl -> MgCl2 + H2

a________2a_______a______a(mol)

MgO +2 HCl -> MgCl2 + H2O

b_____2b_______b___b(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}24a+40b=8,8\\22,4a=4,48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)

=> mMg=0,2.24=4,8(g)

=>%mMg= (4,8/8,8).100=54,545%

=> %mMgO= 45,455%

b) m(muối)=mMg2+ + mCl- = 0,3. 24 + 0,6.35,5=28,5(g)

c) V=VddHCl=(2a+2b)/2=0,3(l)=300(ml)

Câu 2:

Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\n_{HCl}=0,4\cdot2=0,8\left(mol\right)\end{matrix}\right.\)

PTHH: \(Ca+2HCl\rightarrow CaCl_2+H_2\uparrow\)

               0,2____0,4_____0,2____0,2   (mol)

           \(CaO+2HCl\rightarrow CaCl_2+H_2O\)

                0,2____0,4______0,2____0,2  (mol)

Ta có: \(\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{0,2\cdot40}{0,2\cdot40+0,2\cdot56}\cdot100\%\approx41,67\%\\\%m_{CaO}=58,33\%\\m_{CaCl_2}=\left(0,2+0,2\right)\cdot111=44,4\left(g\right)\end{matrix}\right.\)

Ta có: n_N2 = 0,22 (mol)

⇒ n_NO3^- = 10n_N2 = 2,2 (mol)

⇒ m _muối = m_X + m_NO3^- = 31 + 2,2.62 = 167,4 (g)

n_HNO3 = 12n_N2 = 2,64 (mol)

\(\Rightarrow V_{HNO_3}=\dfrac{2,64}{2}=1,32\left(l\right)=1320\left(ml\right)\)

Đổi 100ml=0,1l

\(Al+2HCl\rightarrow AlCl_2+H_2\)

tl1........2.............1..........1.(mol)
Br0,15...0,3......0,15.....0,15(mol)

\(n_{HCl}=C_M.Vdd=0,1.3=0,3\left(mol\right)\)

\(m_{Al}=n.M=0,15.27=4,05\left(g\right)\)

\(V_{H_2}=n.22,4=3,36\left(l\right)\)

\(n_{HCl}=0.1\cdot3=0.3\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(0.1........0.3..........................0.15\)

\(m_{Al}=0.1\cdot27=2.7\left(g\right)\)

\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)