Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
M + 2HCl → MCl2 + H2↑
MO + 2HCl → MCl2 + H2O
MCl2 + 2NaOH → M(OH)2↓ + 2NaCl
M(OH)2 → MO + H2O
M + CuCl2 → MCl2 + Cu↓
a, \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
mhh Zn và Fe = 21,6-3 = 18,6 (g)
PTHH: Zn + H2SO4 → ZnSO4 + H2
Mol: x x
PTHH: Fe + H2SO4 → FeSO4 + H2
Mol: y y
Ta có: \(\left\{{}\begin{matrix}65x+56y=18,6\\x+y=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(\%m_{Zn}=\dfrac{0,2.65.100\%}{21,6}=60,19\%\)
\(\%m_{Fe}=\dfrac{0,1.56.100\%}{21,6}=25,93\%\)
\(\%m_{Cu}=100-60,19-25,93=13,88\%\)
b,
PTHH: Zn + H2SO4 → ZnSO4 + H2
Mol: 0,2 0,2 0,2
PTHH: Fe + H2SO4 → FeSO4 + H2
Mol: 0,1 0,1 0,1
\(m_{H_2SO_4}=\left(0,1+0,2\right).98=29,4\left(g\right)\Rightarrow m_{ddH_2SO_4}=\dfrac{29,4.100\%}{25\%}=117,6\left(g\right)\)
c,mdd sau pư = 21,6+117,6- (0,1+0,2).2 = 138,6 (g)
\(C\%_{ddZnSO_4}=\dfrac{0,2.161.100\%}{138,6}=23,23\%\)
\(C\%_{ddFeSO_4}=\dfrac{0,1.152.100\%}{138,6}=10,97\%\)
a. PTHH : Mg + 2HCl ➝ MgCl2 + H2 (1)
b. theo bài : nH2 = 3,36 : 22,4 = 0,15 (mol)
theo (1) nMg = nH2 = 0,15 (mol)
➞ mMg = 0,15 ✖ 24 = 3,6 (g)
➞ %mMg = (3,6 : 5)✖100 = 72%
➞ %mCu = 100% - 72% = 28%
c. theo (1) nHCl = 2nH2 = 2✖0,15 = 0,3 (mol)
mHCl = 0,3✖36,5 = 10,95(g)
➜mddHCl = (10,95✖100):14,6 = 75(g)
d. dung dịch Y : MgCl2
mdd(spư)= 3,6+75-0,3 = 78,3(g)
theo (1) nMgCl2 = nH2 = 0,15(mol)
mMgCl2 = 0,15✖95 = 14,25(g)
C%MgCl2 = (14,25 : 78,3)✖100 = 18,199%
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: \(m_{H_2}=1\cdot2=2\left(g\right)\)
\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{H_2}=1230\left(g\right)\)
\(\Rightarrow C\%_{MgCl_2}=\dfrac{114}{1230}\cdot100\%\approx9,27\%\)
