Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n_{HCl}=\dfrac{100.36,5\%}{100\%}:36,5=1\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,02-->0,04----->0,02---->0,02
Xét: \(\dfrac{0,02}{1}< \dfrac{1}{2}\) => HCl dư
A gồm \(\left\{{}\begin{matrix}n_{HCl}=1-0,04=0,06\left(mol\right)\\n_{MgCl_2}=0,02\left(mol\right)\end{matrix}\right.\)
\(m_{dd.A}=0,02.24+100-0,02.2=100,44\left(g\right)\)
\(C\%_{HCl}=\dfrac{0,06.36,5.100\%}{100,44}=2,18\%\)
\(C\%_{MgCl_2}=\dfrac{0,02.95.100\%}{100,44}=1,89\%\)
\(m_{HCl}=\dfrac{100.36,5}{100}=36,5\left(g\right)\\ n_{HCl}=\dfrac{36,5}{36,5}=1\left(mol\right)\\ n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\\ CaCO_3+2HCl\xrightarrow[]{}CaCl_2+CO_2+H_2O\\ \dfrac{0,1}{1}< \dfrac{1}{2}\Rightarrow HCl.dư\\ n_{CaCO_3}=n_{CaCl_2}=n_{CO_2}=0,1mol\\ m_{CaCl_2}=0,1.111=11,1\left(g\right)\\m_{CO_2}=0,1.44=4,4\left(g\right)\\ m_{ddCaCl_2}=10+100-4,4=105,6\left(g\right)\\ C_{\%CaCl_2}=\dfrac{11,1}{105,6}\cdot100\%\approx10,5\%\)
Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
a. PTHH: \(Mg+2HCl--->MgCl_2+H_2\)
Theo PT: \(n_{Mg}=n_{H_2}=0,2\left(mol\right)\)
=> \(m_{Mg}=0,2.24=4,8\left(g\right)\)
Theo PT: \(n_{HCl}=2.n_{Mg}=2.0,2=0,4\left(mol\right)\)
=> \(m_{HCl}=0,4.36.5=14,6\left(g\right)\)
=> \(C_{\%_{HCl}}=\dfrac{14,6}{200}.100\%=7,3\%\)
b. Ta có: \(m_{dd_{MgCl_2}}=4,8+200=204,8\left(g\right)\)
Theo PT: \(n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\)
=> \(m_{MgCl_2}=0,2.95=19\left(g\right)\)
=> \(C_{\%_{MgCl_2}}=\dfrac{19}{204,8}.100\%=9,28\%\)
Câu 3 :
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)
1 2 1 1
0,2 0,4 0,2 0,2
b) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
\(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(m_{ddHCl}=\dfrac{14,6.100}{10,95}=133,3\left(g\right)\)
c) \(n_{H2}=\dfrac{0,4.1}{2}=0,2\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)
d) \(n_{MgCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{MgCl2}=0,2.95=19\left(g\right)\)
\(m_{ddspu}=4,8+133,3-\left(0,2.2\right)=137,7\left(g\right)\)
\(C_{MgCl2}=\dfrac{19.100}{137,7}=13,8\)0/0
Chúc bạn học tốt
\(a.m_{HCl}=100.10\%+150.20\%=40\left(g\right)\\ C\%_{ddHCl}=\dfrac{40}{100+150}.100=16\%\\ b.n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{10,95\%.100}{36,5}=0,3\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Vì:\dfrac{0,1}{1}< \dfrac{0,3}{1}\Rightarrow HCldư\\ n_{HCl\left(dư\right)}=0,3-2.0,1=0,1\left(mol\right)\\ m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\\ n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\\ m_{MgCl_2}=0,1.95=9,5\left(g\right)\\ m_{ddsau}=2,4+100-0,1.2=102,2\left(g\right)\\ C\%_{ddHCl\left(dư\right)}=\dfrac{3,65}{102,2}.100\approx3,571\%\)
\(C\%_{ddMgCl_2}=\dfrac{9,5}{102,2}.100\approx9,295\%\)
C% dung dịch sau cùng thì phải có cả dư và tạo thành chứ.
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ m_{HCl}=\dfrac{100.36,5}{100}=36,5\left(g\right)\\ n_{HCl}=\dfrac{36,5}{36,5}=1\left(mol\right)\\ Mg+2HCl\xrightarrow[]{}MgCl_2+H_2\\ \dfrac{0,2}{1}< \dfrac{1}{2}\Rightarrow HCl.dư\\ n_{Mg}=n_{MgCl_2}=n_{H_2}=0,2mol\\ m_{MgCl_2}=0,2.95=19\left(g\right)\\ m_{H_2}=0,2.2=0,4\left(g\right)\\ m_{ddMgCl_2}=4,8+100-0,4=104,4\left(g\right)\\ C_{\%MgCl_2}=\dfrac{19}{104,4}\cdot100\approx18,19\%\)