\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ n_{H_2SO_4}=\dfrac{117,6.25\%}{98}=0,3\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\LTL:\dfrac{0,2}{1}< \dfrac{0,3}{1}\Rightarrow H_2SO_4dư\\ n_{H_2}=n_{Mg}=0,2\left(mol\right)\\ \Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\\DungdịchB:MgSO_4;H_2SO_4dư\\ m_{ddsaupu}=4,8+117,6-0,2.2=122\left(g\right)\\ n_{H_2SO_4dư}=0,3-0,2=0,1\left(mol\right)\\ n_{MgSO_4}=n_{Mg}=0,1\left(mol\right)\\ C\%_{H_2SO_4dư}=\dfrac{0,1.98}{122}.100=8,03\%\\ C\%_{MgSO_4}=\dfrac{0,2.120}{122}.100=19,67\% \)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\) 
          0,1           0,2            0,1      0,1 
\(m_{HCl}=0,2.36,5=7,3\left(g\right)\\ V_{H_2}=0,1.22,4=2,24l\\ m_{\text{dd}}=6,5+200-\left(0,1.2\right)=206,3g\)  
bài 2 :
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\) 
          0,2             0,4       0,2              0,2 
\(m_{HCl}=0,4.36,5=14,6g\\ V_{H_2}=0,2.22,4=4,48l\\ m\text{dd}=4,8+200-0,4=204,4g\\ C\%=\dfrac{0,2.136}{204,4}.100\%=13,3\%\)

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

\(n_{H_2SO_4}=\dfrac{117,6.25\%}{98}=0,3\left(mol\right)\)

PTHH: Mg + H₂SO₄ --> MgSO₄ + H₂

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\) => Mg hết, H₂SO₄ dư

PTHH: Mg + H₂SO₄ --> MgSO₄ + H₂

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

=> V_H2 = 0,2.22,4 = 4,48 (l)

b) m_{dd sau pư} = 4,8 + 117,6 - 0,2.2 = 122 (g)

\(\left\{{}\begin{matrix}C\%_{MgSO_4}=\dfrac{0,2.120}{122}.100\%=19,67\%\\C\%_{H_2SO_4.dư}=\dfrac{\left(0,3-0,2\right).98}{122}.100\%=8,03\%\end{matrix}\right.\)

sao lại tính C% của H2SO4 dư bao nhiêu ? mà k tính toàn bộ

a) PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)

b) Ta có: \(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\) 

\(\Rightarrow\left\{{}\begin{matrix}n_{MgCl_2}=0,2\left(mol\right)\\n_{HCl}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{MgCl_2}=0,2\cdot95=19\left(g\right)\\C\%_{HCl}=\dfrac{0,4\cdot36,5}{200}\cdot100\%=7,3\%\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{MgO}+m_{ddHCl}=208\left(g\right)\)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{19}{208}\cdot100\%\approx9,13\%\)

c) PTHH: \(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_2\downarrow\)

Ta có: \(\left\{{}\begin{matrix}n_{MgCl_2}=0,2\left(mol\right)\\n_{NaOH}=\dfrac{200\cdot4\%}{40}=0,2\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,2}{2}\) \(\Rightarrow\) NaOH p/ứ hết, MgCl₂ còn dư

\(\Rightarrow\left\{{}\begin{matrix}n_{NaCl}=0,2\left(mol\right)\\n_{Mg\left(OH\right)_2}=0,1\left(mol\right)=n_{MgCl_2\left(dư\right)}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,2\cdot58,5=11,7\left(g\right)\\m_{MgCl_2\left(dư\right)}=9,5\left(g\right)\\m_{Mg\left(OH\right)_2}=0,1\cdot58=5,8\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{ddA}+m_{ddNaOH}-m_{Mg\left(OH\right)_2}=402,2\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{11,7}{402,2}\cdot100\%\approx2,91\%\\C\%_{MgCl_2\left(dư\right)}=\dfrac{9,5}{402,2}\cdot100\%\approx2,36\%\end{matrix}\right.\) 

Theo gt ta có: $n_{MgO}=0,2(mol)$

a, $MgO+2HCl\rightarrow MgCl_2+H_2O$

b, Ta có: $n_{HCl}=0,4(mol)\Rightarrow x=7,3$

Bảo toàn khối lượng ta có: $m_{ddA}=208(g)$

$\Rightarrow \%C_{MgCl_2}=9,13\%$

c, Ta có: $n_{NaOH}=0,2(mol)$

$\Rightarrow n_{Mg(OH)_2}=0,1(mol)$

Bảo toàn khối lượng ta có: $m_{ddB}=208+200-0,1.58=402,2(g)$

$\Rightarrow \%C_{MgCl_2}=2,36\%$

a)

$MgO + 2HCl \to MgCl_2  + H_2O$

b)

$n_{MgO} = \dfrac{8}{40} = 0,2(mol)$

$n_{HCl} = 2n_{MgO} = 0,4(mol)$

$C\%_{HCl} = \dfrac{0,4.36,5}{200}.100\% = 7,3\%$

(x = 7,3)

$m_{dd\ A} = 8 + 200 = 208(gam)$
$C\%_{MgCl_2} = \dfrac{0,2.95}{208}.100\% = 9,13\%$

\(a,PTHH:MgO+2HCl\rightarrow MgCl_2+H_2O\)

b, Theo PTHH : \(n_{HCl}=2n_{MgO}=2.\dfrac{m}{M}=0,4\left(mol\right)\)

\(\Rightarrow x=7,3\%\)

Theo PTHH : \(n_{MgCl2}=n_{MgO}=0,2\left(mol\right)\)

\(\Rightarrow m_{MgCl2}=19\left(g\right)\)

Mà mdd = \(m_{MgO}+m_{ddHCl}=208\left(g\right)\)

\(\Rightarrow C\%_{MgCl2}=\dfrac{m}{m_{dd}}.100\%=9,13\%\)

c, \(PTHH:MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2+2NaCl\)

....................0,1..............0,2...............0,1.............0,2.......

Ta có : \(n_{NaOH}=0,2\left(mol\right)\)

=> mdd = \(m_{MgCl2}+m_{NaOH}-m_{Mg\left(OH\right)2}=213,2g\)

- Thấy sau phản ứng dung dịch B gồm NaCl ( 0,2 mol ), MgCl2 dư ( 0,1mol )

\(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=11,7g\\m_{MgCl2}=9,5g\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=5,5\%\\C\%_{MgCl2}=4,46\%\end{matrix}\right.\)

A : $NaOH,NaAlO_2$

B : $Mg$

D : $Al(OH)_3$

E : $NaHCO_3$
F : $Al_2O_3$

$G : MgCl_2$
$H : H_2$

$X : MgCO_3$
$Y : MgO$

Mg + 2HCl \(\rightarrow\) MgCl₂ + H₂

n_Mg = 4,8_/24 = 0,2(mol)

m_HCl = \(\dfrac{m_{dd}.C\%}{100\%}=\dfrac{36,5.20\%}{100\%}=7,3\left(g\right)\)

=> n_HCl = 7,3_/36,5 =0,2(mol)

Lập tỉ lệ :

\(\dfrac{n_{Mg\left(ĐB\right)}}{n_{Mg\left(PT\right)}}=\dfrac{0,2}{1}=0,2\) > \(\dfrac{n_{HCl\left(ĐB\right)}}{n_{HCl\left(PT\right)}}=\dfrac{0,2}{2}=0,1\)

=> Sau phản ứng : Mg dư , HCl hết

=> dd thu được là dd MgCl₂

Theo PT => n_H2 = 1_/2 . n_HCl = 1_/2 . 0,2 = 0,1(mol)

=> m_H2 = 0,1 . 2 = 0,2 (g)

Theo ĐLBTKL : m_{dd sau pứ} = m_Mg + m_{dd HCl} - m_H2 =4,8+ 36,5 - 0,2 =41,1(g)

Theo PT => n_MgCl2 = 1_/2 . n_HCl = 1_/2 . 0,2 =0,1(mol)

=> m_MgCl2 = 0,1 . 95 = 9,5(g)

=> C%_{dd thu được =} 9,5_/41,1 . 100% =23,11%

\(Mg+2HCl(0,2)--->MgCl_2(0,1)+H_2(0,1)\)

\(nMg=0,2(mol)\)

\(mHCl=\dfrac{20.36,5}{100}=7,3\left(g\right)\)

\(\Rightarrow nHCl=0,2\left(mol\right)\)

So sánh \(\dfrac{nMg}{1}=0,2>\dfrac{nHCl}{2}=0,1\)

=> Mg còn dư, chọn nHCl để tính

Theo PTHH: \(nMgCl_2=0,1\left(mol\right)\)

\(\Rightarrow mMgCl_2=9,5\left(g\right)\)

\(nH_2=0,1\left(mol\right)\)

\(\Rightarrow mH_2=0,2\left(g\right)\)

\(m dd sau =4,8+36,5-0,2=41,1(g)\)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{9,5}{41,1}.100\%=23,11\%\)

Ta co pthh

Mg + 2HCl \(\rightarrow\) MgCl2 + H2

Theo de bai ta co

nMg=\(\dfrac{4,8}{24}=0,2mol\)

mHCl=\(\dfrac{mdd.C\%}{100\%}=\dfrac{36,5.20\%}{100\%}=7,3g\Rightarrow nHCl=\dfrac{7,3}{36,5}=0,2mol\)

Theo pthh

nMg=\(\dfrac{0,2}{1}mol>nHCl=\dfrac{0,2}{1}mol\)

\(\Rightarrow\)nMg du ( tinh theo nHCl)

Theo pthh

nMgCl2=\(\dfrac{1}{2}nHCl=\dfrac{1}{2}.0,2=0,1mol\)

\(\Rightarrow\) mMgCl2=0,1.95=9,5 g

Theo pthh

nH2=\(\dfrac{1}{2}nMg=\dfrac{1}{2}.0,2=0,1mol\)

\(\Rightarrow\) mH2=0,1.2=0,2 g

\(\Rightarrow\) mdd= mct + mddHCl - mchatkhi=4,8+36,5-0,2=41,1 g

\(\Rightarrow\) Nong do % chat thu duoc la

C%=\(\dfrac{mct}{mdd}.100\%=\dfrac{9,5}{41,1}.100\%=23,1\%\)