mHCl(trong dd 20%)= 0,2.600=120(g)

Gọi x,y lần lượt là số gam dung dịch HCl 10% và HCl 25% cần lấy để pha trộn (0<x,y<600)

Vì , số gam chất tan tổng là 120 nên ta có pt:

0,1x+0,25y=120 (a)

Mặt khác, dung dịch thu được có khối lượng 600(g) nên ta có pt:

x+y=600(b)

Từ (a),(b) ta được hpt:

\(\left\{{}\begin{matrix}0,1x+0,25y=120\\x+y=600\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=200\\y=400\end{matrix}\right.\)

=> Cần lấy 200g dung dịch HCl 10% và 400g dung dịch HCl 25% để pha trộn thu được 600g dung dịch HCl 20%

a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)

b, \(n_{Mg}=\dfrac{1,2}{24}=0,05\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Mg}=0,1\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,1.36,5}{10\%}=36,5\left(g\right)\)

c, \(n_{H_2}=n_{MgCl_2}=n_{Mg}=0,05\left(mol\right)\)

Ta có: m _{dd sau pư} = 1,2 + 36,5 - 0,05.2 = 37,6 (g)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,05.95}{37,6}.100\%\approx12,63\%\)

PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)

a) Ta có: \(n_{CaCO_3}=\dfrac{2,5}{100}=0,025\left(mol\right)\)

\(\Rightarrow n_{HCl}=0,05mol\) \(\Rightarrow m_{ddHCl}=\dfrac{0,05\cdot36,5}{18\%}\approx10,14\left(g\right)\)

b) Theo PTHH: \(n_{CaCl_2}=n_{CO_2}=n_{CaCO_3}=0,025\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CaCl_2}=0,025\cdot111=2,775\left(g\right)\\m_{CO_2}=0,025\cdot44=1,1\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(saup/ứ\right)}=m_{Zn}+m_{ddHCl}-m_{CO_2}=11,54\left(g\right)\)

\(\Rightarrow C\%_{CaCl_2}=\dfrac{2,775}{11,54}\cdot100\%\approx24,05\%\)

CaCO₃+2HCl→CaCl₂+CO₂↑ +H₂O

\(+n_{CaCO_3}=\dfrac{2,5}{100}=0,025\left(mol\right)\)

\(+n_{HCl}=2n_{CaCO_3}=0,05\left(mol\right)\)

\(+m_{HCl}=0,05.98=4,9\left(gam\right)\)

\(+m_{dungdịchHCl}=\dfrac{4,9}{18}.100\%=27,2\left(gam\right)\)

\(+n_{CaCl}=n_{CaCO_3}=0,025\left(mol\right)\)

\(+m_{CaCl_2}=0,025.111=2,775\left(gam\right)\)

Theo ĐLBTKL ta có:

\(m_{CaCl_2}=2,5+27,2-0,025.44-0,025.18=28,15\left(gam\right)\)

C%=\(\dfrac{2,775}{28,15}.100\%\approx9,85\%\)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\\ 2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow n_{Al}=0,2(mol)\\ \Rightarrow m_{Al}=0,2.27=5,4(g)\\ \Rightarrow \%_{Al}=\dfrac{5,4}{10}.100\%=54\%\\ \Rightarrow \%_{Ag}=100\%-54\%=46\%\\ n_{HCl}=0,6(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,6.36,5}{10\%}=219(g)\)

Fe + 2HCl -->. FeCl2 + H2

0,15-----0,3
Cu không pứ với HCl
nH2 =3,36\22,4= 0,15 mol
=> nFe = 0,15 mol
=>mFe = 0,15.56 = 8,4 g
=>mCu=10-8,4=1,6g

=>mHCl=0,3.98=29,4g

=>mddHCl=80,55g

a) 

\(m_{H_2SO_4}=\dfrac{300.19,6}{100}=58,8\left(g\right)\)

=> \(m_{dd.H_2SO_4.9,8\%}=\dfrac{58,8.100}{9,8}=600\left(g\right)\)

=> \(m_{H_2O\left(thêm\right)}=600-300=300\left(g\right)\)

b)

\(n_{HCl}=0,2.2=0,4\left(mol\right)\)

=> \(V_{dd.HCl.1,5M}=\dfrac{0,4}{1,5}=\dfrac{4}{15}\left(l\right)\)

=> \(V_{H_2O\left(thêm\right)}=\dfrac{4}{15}-0,2=\dfrac{1}{15}\left(l\right)=\dfrac{200}{3}\left(ml\right)\)

=> \(m_{H_2O\left(thêm\right)}=\dfrac{200}{3}.1=\dfrac{200}{3}\left(g\right)\)

\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(0.2.......0.2......................0.2\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(C\%H_2SO_4=\dfrac{0.2\cdot98\cdot100\%}{200}=9.8\%\)

\(n_{Mg}=\dfrac{2.4}{24}=0.1\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(0.1.......0.2...........0.1........0.1\)

\(m_{dd_{HCl}}=\dfrac{0.2\cdot36.5\cdot100}{14.6}=50\left(g\right)\)

\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(m_{MgCl_2}=0.1\cdot95=9.5\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng}}=2.4+50-0.1\cdot2=52.2\left(g\right)\)

\(C\%_{MgCl_2}=\dfrac{9.5}{52.2}\cdot100\%=18.2\%\)

a/ \(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)

PTHH: Mg + 2HCl → MgCl₂ + H₂

Mol:     0,3      0,6         0,3       0,3

\(m_{Mg}=0,3.24=7,2\left(g\right)\)

b/ \(m_{MgCl_2}=0,3.95=28,5\left(g\right)\)

c/ \(m_{HCl}=0,3.36,5=10,95\left(g\right)\)