Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
nH2 = \(\frac{1,68}{22,4}\) = 0,075 (mol)
Mg + 2HCl \(\rightarrow\) MgCl2 + H2\(\uparrow\) (1)
0,075 <--------0,075 <--0,075 (mol)
MgO + 2HCl \(\rightarrow\) MgCl2 + H2O (2)
%mMg= \(\frac{0,075.24}{5,8}\) . 100% = 31,03 %
%m MgO = 68,97%
nMgO = \(\frac{5,8-0,075.24}{40}\) = 0,1 (mol)
Theo pt(2) nMgCl2 = nMgO= 0,1 (mol)
mdd sau pư = 5,8 + 194,35 - 0,075.2 = 200 (g)
C%(MgCl2) = \(\frac{95\left(0,075+0,1\right)}{200}\) . 100% = 8,3125%
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!
\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ a,PTHH:MgCO_3+H_2SO_4\rightarrow MgSO_4+H_2O+CO_2\uparrow\\ MgO+H_2SO_4\rightarrow MgSO_4+H_2O\\ \Rightarrow n_{MgCO_3}=n_{CO_2}=0,1\left(mol\right)\\ \Rightarrow m_{MgCO_3}=0,1\cdot84=8,4\left(g\right)\\ \Rightarrow\%_{MgCO_3}=\dfrac{8,4}{10,4}\cdot100\%\approx80,77\%\\ \Rightarrow\%_{MgO}=100\%-80,77\%=19,23\%\)
\(b,m_{MgO}=10,4-8,4=2\left(g\right)\\ \Rightarrow n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\\ \Rightarrow\sum n_{H_2SO_4}=n_{MgCO_3}+n_{MgO}=0,15\left(mol\right)\\ \Rightarrow\sum m_{H_2SO_4}=0,15\cdot98=14,7\left(g\right)\\ \Rightarrow\sum m_{dd_{H_2SO_4}}=\dfrac{14,7}{9,8\%}=150\left(g\right)\\ \sum n_{MgSO_4}=\sum n_{H_2SO_4}=0,15\left(mol\right)\\ \Rightarrow\sum m_{MgSO_4}=0,15\cdot120=18\left(g\right)\\ \Rightarrow C\%_{MgSO_4}=\dfrac{18}{10,4+150-0,1\cdot44}\approx11,54\%\)
Ta có: \(n_{SO_3}=\dfrac{8}{80}=0,1\left(mol\right)\)
a. PTHH: SO3 + H2O ---> H2SO4 (1)
b. Theo PT(1): \(n_{H_2SO_4}=n_{SO_3}=0,1\left(mol\right)\)
Đổi 250ml = 0,25 lít
=> \(C_{M_{H_2SO_4}}=\dfrac{0,1}{0,25}=0,4\left(M\right)\)
c. PTHH: H2SO4 + 2KOH ---> K2SO4 + 2H2O
Theo PT(2): \(n_{KOH}=2.n_{H_2SO_4}=2.0,1=0,2\left(mol\right)\)
=> \(m_{KOH}=0,2.56=11,2\left(g\right)\)
Ta có: \(C_{\%_{KOH}}=\dfrac{11,2}{m_{dd_{KOH}}}.100\%=5,6\%\)
=> \(m_{dd_{KOH}}=200\left(g\right)\)
Ta có: \(d_{KOH}=\dfrac{200}{V_{dd_{KOH}}}=1,045\)(g/ml)
=> \(V_{dd_{KOH}}=191,4\left(ml\right)\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,3\left(mol\right)=n_{ZnCl_2}\\n_{HCl}=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0.3\cdot65}{35,5}\cdot100\%\approx54,93\%\\\%m_{Cu}=45,07\%\\C\%_{HCl}=\dfrac{0,6\cdot36,5}{500}\cdot100\%=4,38\%\\m_{ZnCl_2}=0,3\cdot136=40,8\left(g\right)\end{matrix}\right.\)
Mặt khác: \(\left\{{}\begin{matrix}m_{Cu}=35,5-0,3\cdot65=16\left(g\right)\\m_{H_2}=0,3\cdot2=0,6\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m_{dd}=m_{KL}+m_{ddHCl}-m_{Cu}-m_{H_2}=518,9\left(g\right)\)
\(\Rightarrow C\%_{ZnCl_2}=\dfrac{40,8}{518,9}\cdot100\%\approx7,86\%\)
\(a.Mg+2HCl->MgCl_2+H_2\\ MgO+2HCl->MgCl_2+H_2O\\ b.n_{H_2}=\dfrac{2,24}{22,4}=n_{Mg}=0,1mol\\ \%m_{Mg}=\dfrac{0,1.24}{6}=40\%;\%m_{MgO}=60\%\\ n_{MgO}=\dfrac{0,6.6}{40}=0,09\left(mol\right)\\ n_{MgCl_2}=0,1+0,09=0,19\left(mol\right)\\ n_{HCl}=0,19.2=0,38\left(mol\right)\\ V_{ddHCl}=\dfrac{0,38.36,5}{0,2.1,1}=63,0\left(mL\right)\\ C_{M\left(MgCl_2\right)}=\dfrac{0,19}{0,063}=3,0\left(M\right)\)
Mình
hk chắc nha 😅😅😅😅