Ta có: \(n_{MgO}=\dfrac{10}{40}=0,25\left(mol\right)\)

PTHH: MgO + 2HCl ---> MgCl₂ + H₂O

Ta có: \(m_{dd_{MgCl_2}}=10+115=125\left(g\right)\)

Theo PT: \(n_{MgCl_2}=n_{MgO}=0,25\left(mol\right)\)

=> \(m_{MgCl_2}=0,25.95=23,75\left(g\right)\)
=> \(C_{\%_{MgCl_2}}=\dfrac{23,75}{125}.100\%=19\%\)

sao em cứ cộng mH2O vào tính mdd nhỉ?

\(n_{MgO}=\frac{10}{40}=0,25\left(mol\right)\)

\(m_{HCl}=\frac{115.26\%}{100\%}=29,9\left(g\right)\)

\(n_{HCl}=\frac{29,9}{36,5}\approx0,82\left(mol\right)\)

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

Ban đầu: \(0,25\)_____\(0,82\)

Phản ứng: \(0,25\)____\(0,5\)____\(0,25\)______\(0,25\) \(\left(mol\right)\)

Dư:______________\(0,32\)

Lập tỉ lệ: \(\frac{0,25}{1}< \frac{0,82}{2}\left(0,25< 0,41\right)\)

\(\Rightarrow MgO\) hết \(HCl\) dư

Các chất sau phả ứng là \(HCl\left(dư\right)\) và \(MgCl_2\)

\(m_{H_2O}=0,25.18=4,5\left(g\right)\)

\(m_{ddsaupư}=m_{MgO}+m_{ddHCl}+m_{H_2O}=10+115+4,5=129,5\left(g\right)\)

\(m_{HCl\left(dư\right)}=0,32.36,5=11,68\left(g\right)\)

\(C\%_{HCl\left(dư\right)}=\frac{11,68}{129,5}.100\%=9\%\)

\(m_{MgCl_2}=0,25.95=23,75\left(g\right)\)

\(C\%_{MgCl_2}=\frac{23,75}{129,5}.100\%=18,34\%\)

$n_{MgO} = \dfrac{6}{40} = 0,15(mol)$
$n_{HCl} = 125.14,6\% : 36,5 = 0,5(mol)$
$MgO + 2HCl \to MgCl_2 + H_2O$

Ta thấy : 

$n_{MgO} : 1 = 0,15 < n_{HCl} : 2 = 0,25$ nên HCl dư

$n_{HCl\ pư} = 2n_{MgO} = 0,3(mol)$

$n_{MgCl_2} = n_{MgO} = 0,15(mol)$

Sau phản ứng : 

$m_{dd} = 6 + 125 = 131(gam)$
Vậy : 

$C\%_{MgCl_2} = \dfrac{0,15.95}{131}.100\% = 10,88\%$

$C\%_{HCl} = \dfrac{(0,5 - 0,3).36,5}{131}.100\% = 5,6\%$

\(n_{MgO}=\dfrac{6}{40}=0.15\left(mol\right)\)

\(n_{HCl}=\dfrac{125\cdot14.6\%}{36.5}=0.5\left(mol\right)\)

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

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

\(m_{\text{dung dịch sau phản ứng}}=6+125=131\left(g\right)\)

\(C\%_{MgCl_2}=\dfrac{0.15\cdot95}{131}\cdot100\%=10.8\%\)

\(C\%_{HCl\left(dư\right)}=\dfrac{\left(0.5-0.3\right)\cdot35.5}{131}\cdot100\%=5.42\%\%\)

nMgO=0,15(mol); nH2SO4=0,4(mol)

PTHH: MgO + H2SO4 -> MgSO4 + H2O

0,15________0,15__________0,15(mol)

Ta có: 0,15/1 < 0,4/1

=> H2SO4 dư, MgO hết, tính theo nMgO

-> nH2SO4(dư)=0,4-0,15=0,25(mol) => mH2SO4(dư)=24,5(g)

nMgSO4=nMg=0,15(mol) => mMgSO4=120.0,15=18(g)

mddsau=6+200=206(g)

=>C%ddH2SO4(dư)=(24,5/206).100=11,893%

C%ddMgSO4=(18/206).100=8,738%

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

Ta có: \(\left\{{}\begin{matrix}n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\\n_{H_2SO_4}=\dfrac{200\cdot19,6\%}{98}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Axit dư

\(\Rightarrow\left\{{}\begin{matrix}n_{MgSO_4}=n_{H_2}=0,25\left(mol\right)\\n_{H_2SO_4\left(dư\right)}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{MgSO_4}=0,25\cdot120=30\left(g\right)\\m_{H_2}=0,25\cdot2=0,5\left(g\right)\\m_{H_2SO_4\left(dư\right)}=0,15\cdot98=14,7\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Mg}+m_{ddH_2SO_4}-m_{H_2}=205,5\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgSO_4}=\dfrac{30}{205,5}\cdot100\%\approx14,6\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{14,7}{205,5}\cdot100\%\approx7,2\%\end{matrix}\right.\)

PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)  (1)

             \(MgO+2HCl\rightarrow MgCl_2+H_2O\)  (2)

a) Ta có: \(n_{H_2}=\dfrac{22,4}{22,4}=1\left(mol\right)=n_{Mg}\) \(\Rightarrow m_{Mg}=1\cdot24=24\left(g\right)\)

\(\Rightarrow\%m_{Mg}=\dfrac{24}{32}\cdot100\%=75\%\) \(\Rightarrow\%m_{MgO}=25\%\)

b) Theo 2 PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(1\right)}=2n_{Mg}=2mol\\n_{HCl\left(2\right)}=2n_{MgO}=2\cdot\dfrac{32-24}{40}=0,4mol\end{matrix}\right.\)

\(\Rightarrow\Sigma n_{HCl}=2,4mol\) \(\Rightarrow m_{ddHCl}=\dfrac{2,4\cdot36,5}{7,3\%}=1200\left(g\right)\)

c) Theo PTHH: \(\Sigma n_{MgCl_2}=\dfrac{1}{2}\Sigma n_{HCl}=1,2mol\)

\(\Rightarrow\Sigma m_{MgCl_2}=1,2\cdot95=114\left(g\right)\)

Mặt khác: \(m_{H_2}=1\cdot2=2\left(g\right)\)

\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{H_2}=1230\left(g\right)\)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{114}{1230}\cdot100\%\approx9,27\%\)

Ta có: \(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)

\(n_{HCl}=4.100:1000=0,4\left(mol\right)\)

a. PTHH: MgO + 2HCl ---> MgCl₂ + H₂O

Ta thấy: \(\dfrac{0,1}{1}< \dfrac{0,4}{2}\)

Vậy HCl dư.

=> \(n_{dư}=\dfrac{0,1.2}{0,4}=0,5\left(mol\right)\)

=> \(m_{dư}=0,5.36,5=18,2\left(g\right)\) 

b. Ta có: \(V_{dd_{MgCl_2}}=V_{HCl}=\dfrac{100}{1000}=0,1\left(lít\right)\)

Theo PT: \(n_{MgCl_2}=n_{MgO}=0,1\left(mol\right)\)

=> \(C_{M_{MgCl_2}}=\dfrac{0,1}{0,1}=1M\)

cảm ơn bạn nhìu

\(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)

\(n_{H2SO4}=\dfrac{19,6\%.100}{100\%.98}=0,2\left(mol\right)\)

Pt : \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

Xét tỉ lệ : \(\dfrac{0,05}{1}< \dfrac{0,2}{1}\Rightarrow H_2SO_4dư\)

Theo pt : \(n_{MgO\left(pư\right)}=n_{MgSO4}=0,05\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgSO4}=\dfrac{0,05.120}{2+100}.100\%=5,88\%\\C\%_{ddH2SO4\left(dư\right)}=\dfrac{\left(0,2-0,05\right).98}{2+100}.100\%=14,41\%\end{matrix}\right.\)