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nMgO=1040=0,25(mol)nMgO=1040=0,25(mol)
mHCl=115.26%100%=29,9(g)mHCl=115.26%100%=29,9(g)
nHCl=29,936,5≈0,82(mol)nHCl=29,936,5≈0,82(mol)
PTHH:MgO+2HCl→MgCl2+H2OPTHH:MgO+2HCl→MgCl2+H2O
Ban đầu: 0,250,25_____0,820,82
Phản ứng: 0,250,25____0,50,5____0,250,25______0,250,25 (mol)(mol)
Dư:______________0,320,32
Lập tỉ lệ: 0,251<0,822(0,25<0,41)0,251<0,822(0,25<0,41)
⇒MgO⇒MgO hết HClHCl dư
Các chất sau phả ứng là HCl(dư)HCl(dư) và MgCl2MgCl2
mH2O=0,25.18=4,5(g)mH2O=0,25.18=4,5(g)
mddsaupư=mMgO+mddHCl+mH2O=10+115+4,5=129,5(g)mddsaupư=mMgO+mddHCl+mH2O=10+115+4,5=129,5(g)
mHCl(dư)=0,32.36,5=11,68(g)mHCl(dư)=0,32.36,5=11,68(g)
C%HCl(dư)=11,68129,5.100%=9%C%HCl(dư)=11,68129,5.100%=9%
mMgCl2=0,25.95=23,75(g)mMgCl2=0,25.95=23,75(g)
C%MgCl2=23,75129,5.100%=18,34%
- NẾU ĐÚNG ĐÁNH GIÁ CHO MÌNH NHÉ -
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)
\(Ca+2HCl\rightarrow CaCl_2+H_2\)
0,1 0,1 ( mol )
\(\rightarrow\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{0,1.40}{10}.100=40\%\\\%m_{MgO}=100\%-40\%=60\%\end{matrix}\right.\)
\(\left\{{}\begin{matrix}C\%_{CaCl_2}=\dfrac{0,1.111}{10+390,2-0,1.2}.100=2,775\%\\C\%_{MgO}=\dfrac{4}{10+390,2-0,1.2}.100=1\%\end{matrix}\right.\)
$n_{MgO} = \dfrac{6}{40} = 0,15(mol)$
$n_{HCl} = 125.14,6\% : 36,5 = 0,5(mol)$
$MgO + 2HCl \to MgCl_2 + H_2O$
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 :
$C\%_{MgCl_2} = \dfrac{0,15.95}{131}.100\% = 10,88\%$
$C\%_{HCl} = \dfrac{(0,5 - 0,3).36,5}{131}.100\% = 5,6\%$
\(n_{MgO}=\dfrac{6}{40}=0.15\left(mol\right)\)
\(n_{HCl}=\dfrac{125\cdot14.6\%}{36.5}=0.5\left(mol\right)\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
\(0.15...........0.3.......0.15\)
\(m_{\text{dung dịch sau phản ứng}}=6+125=131\left(g\right)\)
\(C\%_{MgCl_2}=\dfrac{0.15\cdot95}{131}\cdot100\%=10.8\%\)
\(C\%_{HCl\left(dư\right)}=\dfrac{\left(0.5-0.3\right)\cdot35.5}{131}\cdot100\%=5.42\%\%\)
\(n_{MgO}=\frac{10}{40}=0,25\left(mol\right)\)
\(m_{HCl}=\frac{115.26\%}{100\%}=29,9\left(g\right)\)
\(n_{HCl}=\frac{29,9}{36,5}\approx0,82\left(mol\right)\)
\(PTHH:MgO+2HCl\rightarrow MgCl_2+H_2O\)
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 \(MgCl_2\)
\(m_{H_2O}=0,25.18=4,5\left(g\right)\)
\(m_{ddsaupư}=m_{MgO}+m_{ddHCl}+m_{H_2O}=10+115+4,5=129,5\left(g\right)\)
\(m_{HCl\left(dư\right)}=0,32.36,5=11,68\left(g\right)\)
\(C\%_{HCl\left(dư\right)}=\frac{11,68}{129,5}.100\%=9\%\)
\(m_{MgCl_2}=0,25.95=23,75\left(g\right)\)
\(C\%_{MgCl_2}=\frac{23,75}{129,5}.100\%=18,34\%\)
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: \(m_{H_2}=1\cdot2=2\left(g\right)\)
\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{H_2}=1230\left(g\right)\)
\(\Rightarrow C\%_{MgCl_2}=\dfrac{114}{1230}\cdot100\%\approx9,27\%\)
a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
b, Giả sử: \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{MgO}=y\left(mol\right)\end{matrix}\right.\)
⇒ 24x + 40y = 4,4 (1)
Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Mg}=x\left(mol\right)\)
⇒ x = 0,1 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1.24}{4,4}.100\%\approx54,54\%\\\%m_{MgO}\approx45,46\%\end{matrix}\right.\)
c, Theo PT: \(\Sigma n_{HCl}=2n_{Mg}+2n_{MgO}=0,3\left(mol\right)\)
\(\Rightarrow V_{HCl}=\dfrac{0,3}{2}=0,15\left(l\right)=150\left(ml\right)\)
Bạn tham khảo nhé!
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: \(n_{MgCl_2}=n_{MgO}=0,25\left(mol\right)\)
=> \(m_{MgCl_2}=0,25.95=23,75\left(g\right)\)
=> \(C_{\%_{MgCl_2}}=\dfrac{23,75}{125}.100\%=19\%\)