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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: \(n_{MgCl_2}=n_{MgO}=0,25\left(mol\right)\)
=> \(m_{MgCl_2}=0,25.95=23,75\left(g\right)\)
=> \(C_{\%_{MgCl_2}}=\dfrac{23,75}{125}.100\%=19\%\)
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 )
\(\rightarrow\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{0,1.40}{10}.100=40\%\\\%m_{MgO}=100\%-40\%=60\%\end{matrix}\right.\)
\(\left\{{}\begin{matrix}C\%_{CaCl_2}=\dfrac{0,1.111}{10+390,2-0,1.2}.100=2,775\%\\C\%_{MgO}=\dfrac{4}{10+390,2-0,1.2}.100=1\%\end{matrix}\right.\)
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: \(n_{MgCl_2}=n_{MgO}=0,1\left(mol\right)\)
=> \(C_{M_{MgCl_2}}=\dfrac{0,1}{0,1}=1M\)
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: \(m_{dd}=m_{Mg}+m_{ddH_2SO_4}-m_{H_2}=205,5\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgSO_4}=\dfrac{30}{205,5}\cdot100\%\approx14,6\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{14,7}{205,5}\cdot100\%\approx7,2\%\end{matrix}\right.\)
\(n_{Fe_2O_3}=\dfrac{10}{160}=0,0625\left(mol\right)\\ n_{HCl}=\dfrac{200.18,25\%}{36,5}=1\left(mol\right)\\ Fe_2O_3+6HCl\xrightarrow[]{}2FeCl_3+3H_2O\\ TC:\dfrac{0,0625}{1}< \dfrac{1}{6}\Rightarrow HCl.dư\\ n_{FeCl_3}=0,0625.2=0,125\left(mol\right)\\ n_{HCl}=0,0625.6=0,375\left(mol\right)\\ m_{dd}=10+200=210\left(g\right)\\ C_{\%FeCl_3}=\dfrac{0,125.162,5}{210}\cdot100\approx9,67\%\\ C_{\%HCl\left(dư\right)}=\dfrac{\left(1-0,375\right).36,5}{210}\cdot100\approx10,86\%\)
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 \(MgCl_2\)
\(m_{H_2O}=0,25.18=4,5\left(g\right)\)
\(m_{ddsaupư}=m_{MgO}+m_{ddHCl}+m_{H_2O}=10+115+4,5=129,5\left(g\right)\)
\(m_{HCl\left(dư\right)}=0,32.36,5=11,68\left(g\right)\)
\(C\%_{HCl\left(dư\right)}=\frac{11,68}{129,5}.100\%=9\%\)
\(m_{MgCl_2}=0,25.95=23,75\left(g\right)\)
\(C\%_{MgCl_2}=\frac{23,75}{129,5}.100\%=18,34\%\)