PTHH:

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

x/2____x

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

y_____y

\(n_{Mg}=\frac{4,8}{24}=0,2\left(mol\right)\)

Gọi x là \(n_{HCl}\)=> V=2000x

y là \(n_{H_2SO_4}\)=>V=4000y

Ta có hệ phương trình: \(\begin{cases}\frac{x}{2}+y=0,2\\2000x=4000y\end{cases}\Leftrightarrow\begin{cases}x=0,2\\y=0,1\end{cases}\)

Vậy V=2.0,2=0,4 (lít)

\(n_{Mg}=\dfrac{4.8}{24}=0.2\left(mol\right)\)

\(Mg+2H^+\rightarrow Mg^{2+}+H_2\)

\(0.2.......0.4....................0.2\)

\(V_{dd}=\dfrac{0.4}{1.5+0.5\cdot2}=0.16\left(l\right)\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

Gọi x là số mol HCl và y là số mol H2SO4

a/ Ta có : \(n_{Mg}=\frac{4,8}{24}=0,2\left(mol\right)\)

PTHH :                      \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

(mol)                          x/2        x                           x/2

                                    \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)

 (mol)                          y              y                            y

Ta có : \(m_{Mg}=24\left(\frac{x}{2}+y\right)=4,8\Rightarrow\frac{x}{2}+y=0,2\Rightarrow x+2y=0,4\)

Mà : \(V_{hh}=V_{HCl}+V_{H_2SO_4}=\frac{x}{1}+\frac{y}{0,5}=x+2y\)

\(\Rightarrow V_{hh}=0,4\left(l\right)\)

b/ Ta có \(n_{H_2}=\frac{x}{2}+y=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,2\times22,4=4,48\left(l\right)\)

a) \(n_{Na}=\dfrac{11,5}{23}=0,5\left(mol\right)\)

\(n_{NaOH}=\dfrac{8\%.500}{40}=1\left(mol\right)\)

\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)

0,5---------------->0,5------->0,25

\(\Sigma n_{NaOH}=0,5+1=1,5\left(mol\right)\)

\(m_{ddsaupu}=11,5+500-0,25.2=511\left(g\right)\)

=> \(C\%_{NaOH}=\dfrac{1,5.40}{511}.100=11,74\%\)

b) Gọi thể tích dung dịch X cần tìm là V

 \(n_{H^+}=V.1+V.0,5.1=2V\left(mol\right)\)

\(H^++OH^-\rightarrow H_2O\)

Ta có : \(n_{H^+}=n_{OH^-}=1,5\left(mol\right)\)

=> 2V=1,5

=> V=0,75(lít)

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

số mol H₂SO₄ :

n_H2SO4 = 0,5 . 0,7 = 0,35 (mol)

theo phương trình hóa học ta có:

n_KOH = 2.n_H2SO4 = 2.0,35 =0,7 (mol)

=> m_KOH = 0,7.56 = 39,2 (g)

=> m_{dd KOH} = (39,2.100)/12 =326,67 (g)

ta có :

V_{dd KOH} = 326,67/1,15 = 284 (ml)

$n_{Mg}=\dfrac{4,8}{24}=0,2mol$

$PTHH :$

$Mg+H_2SO_4\to MgSO_4+H_2$\(\uparrow\)

$Theo\ pt :$

$n_{H_2SO_4}=n_{H_2}=n_{Mg}=0,2mol$

\(\Rightarrow\)$V=V_{dd\ H_2SO_4}=\dfrac{0,2}{0,5}=0,4l$

$V_{H_2}=0,2.22,4=4,48l$

\(n_{HCl}=0,025.0,2=0,005\left(mol\right)\)

PT: \(HCl+NaOH\rightarrow NaCl+H_2O\)

\(n_{NaOH}=n_{HCl}=0,005\left(mol\right)\)

\(\Rightarrow V_{NaOH}=\dfrac{0,005}{0,5}=0,01\left(l\right)=10\left(ml\right)\)

a, \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

Gọi: V_{hh axit} = a (l)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=1,5a\left(mol\right)\\n_{H_2SO_4}=0,5a\left(mol\right)\end{matrix}\right.\)

Theo PT: \(n_{Mg}=\dfrac{1}{2}n_{HCl}+n_{H_2SO_4}\) \(\Rightarrow0,2=\dfrac{1}{2}.1,5a+0,5a\)

⇒ a = 0,16 (l) = 160 (ml)

b, Theo PT: \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)