Cho 15,9g Na2CO3 tác dụng hoàn toàn trong 200g dung dịch HCl
a)Viết phương trình hóa học
b)Tính nồng độ % dung dịch HCl cần dùng?
c)Tính thể tích khí sinh ra ở đkc?
d)Tính nồng độ % của chất trong dung dịch sau khi phản ứng kết thúc
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(Pt:2Al+6HCl\rightarrow2AlCl_3+3H_2\)
b) \(n_{Al}=\dfrac{5,4}{27}=0,2mol\)
\(Theopt:n_{H_2}=\dfrac{3}{2}n_{Al}=0,3mol\)
\(\Rightarrow V_{H_2}=0,3.22,4=6,72lít\)
c) \(Theopt:n_{HCl}=3n_{Al}=0,6mol\)
\(\Rightarrow C_Mdd_{HCl}=\dfrac{0,6}{0,2}=3M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2 0,4 0,2
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\\
C\%_{HCl}=\dfrac{0,4.36,5}{200}.100\%=7,3\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
a) PTHH : \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b) \(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\Rightarrow C_{MddHCl}=\dfrac{0,4}{0,2}=2M\)
c) \(n_{H2}=n_{Zn}=0,2\left(mol\right)\Rightarrow V_{H2\left(dkc\right)}=0,2.24,79=4,958\left(l\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,PTHH:Fe+CuSO_4\rightarrow FeSO_4+Cu\\ b,n_{Fe}=\dfrac{1,96}{56}=0,035\left(mol\right)\\ \Rightarrow n_{CuSO_4}=0,035\left(mol\right)\\ \Rightarrow m_{CT_{CuSO_4}}=0,035\cdot160=5,6\left(g\right)\\ \Rightarrow m_{dd_{CuSO_4}}=\dfrac{5,6\cdot100\%}{10\%}=56\left(g\right)\\ c,n_{FeSO_4}=n_{Cu}=n_{Fe}=0,035\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}m_{CT_{FeSO_4}}=0,035\cdot152=5,32\left(g\right)\\m_{Cu}=0,035\cdot64=2,24\left(g\right)\end{matrix}\right.\\ \Rightarrow m_{dd_{FeSO_4}}=1,96+56-2,24=55,72\left(g\right)\\ \Rightarrow C\%_{FeSO_4}=\dfrac{5,32}{55,72}\cdot100\%\approx9,55\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
b, \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,45\left(mol\right)\Rightarrow V_{H_2}=0,45.22,4=10,08\left(l\right)\)
c, Ta có: m dd sau pư = 8,1 + 200 - 0,45.2 = 207,2 (g)
Theo PT: \(n_{AlCl_3}=n_{Al}=0,3\left(mol\right)\)
\(\Rightarrow C\%_{AlCl_3}=\dfrac{0,3.133,5}{207,2}.100\%\approx19,33\%\)
\(a,PTHH:Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\uparrow\\ b,n_{Na_2CO_3}=\dfrac{15,9}{106}=0,15\left(mol\right)\\ \Rightarrow n_{HCl}=0,3\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,3\cdot36,5=10,95\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{10,95}{200}\cdot100\%=5,475\%\\ c,n_{CO_2}=0,15\left(mol\right)\\ \Rightarrow V_{CO_2\left(đkc\right)}=0,15\cdot24,79=3,7185\left(l\right)\\ d,m_{CO_2}=0,15\cdot44=6,6\left(g\right)\\ n_{NaCl}=0,3\left(mol\right);n_{H_2O}=0,15\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}m_{CT_{NaCl}}=0,3\cdot58,5=17,55\left(g\right)\\m_{H_2O}=0,15\cdot18=2,7\left(g\right)\end{matrix}\right.\\ m_{dd_{NaCl}}=15,9+200-2,7-6,6=206,6\left(g\right)\\ \Rightarrow C\%_{NaCl}=\dfrac{17,55}{206,6}\cdot100\%\approx8,49\%\)