\(n_{CuSO_4}=0.2\cdot0.5=0.1\left(mol\right)\)

\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

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

\(C_{M_{Na_2SO_4}}=\dfrac{0.1}{0.3+0.2}=0.2\left(M\right)\)

\(Cu\left(OH\right)_2\underrightarrow{^{^{t^0}}}CuO+H_2O\)

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

\(m_{CuO}=0.1\cdot80=8\left(g\right)\)

Ta có: \(n_{MgO}=\dfrac{40}{40}=1\left(mol\right)\)

\(n_{HCl}=0,5.1=0,5\left(mol\right)\)

PT: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)

Xét tỉ lệ: \(\dfrac{1}{1}< \dfrac{0,5}{2}\), ta được MgO dư.

Theo PT: \(n_{MgCl_2}=\dfrac{1}{2}n_{HCl}=0,25\left(mol\right)\)

\(\Rightarrow C_{M_{MgCl_2}}=\dfrac{0,25}{0,5}=0,5\left(M\right)\)

PTHH: \(CaCl_2+Na_2CO_3\rightarrow CaCO_3\downarrow+2NaCl\)

a+b) Ta có: \(n_{CaCl_2}=0,1\cdot2=0,2\left(mol\right)\) 

\(\Rightarrow\left\{{}\begin{matrix}n_{CaCO_3}=0,2\left(mol\right)\\n_{NaCl}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaCO_3}=0,2\cdot100=20\left(g\right)\\C_{M_{NaCl}}=\dfrac{0,4}{0,1+0,2}\approx1,33\left(M\right)\end{matrix}\right.\)

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

Theo PTHH: \(n_{HCl}=2n_{CaCO_3}=0,4\left(mol\right)\) \(\Rightarrow m_{ddHCl}=\dfrac{0,4\cdot36,5}{10\%}=146\left(g\right)\)

a.PTHH:CaCl2+Na2CO3--->CaCO3+2NaCl

Ta có:nCaCl2=0,2

=>nCaCO3=nCaCl2=0,2(mol)=>mCaCO3(kết tủa)=100.0,2=2(g)

b.Vdd=100+200=300(ml)=0,3(l)

CM Nacl=(2.0,2)/0,3=4/3(M)(Đề cho 2 chất td vừa đủ nên dd sau pứ chỉ có NaCl)

c.CaCO3+2HCl--->CaCl2+CO2+H2O

nHCl(cần dùng)=2.0.2=0,4(mol)=>mHCl=36,5.0,4=14,6(g)

=>mddHCl=14,6/10%=146(g)

\(a,2NaOH+MgSO_4\rightarrow Mg\left(OH\right)_2+Na_2SO_4\\ n_{NaOH}=0,5.1=0,5\left(mol\right)\\ b,n_{Mg\left(OH\right)_2}=\dfrac{0,5}{2}=0,25\left(mol\right)=n_{Na_2SO_4}\\ m_{kt}=m_{Mg\left(OH\right)_2}=58.0,25=14,5\left(g\right)\\ c,V_{ddX}=V_{ddNaOH}+V_{ddMgSO_4}=0,5+0,5=1\left(l\right)\\ C_{MddNa_2SO_4}=\dfrac{0,25}{1}=0,25\left(M\right)\)

a, Hiện tượng: có kết tủa trắng.

\(n_{KCl}=0,4\cdot2=0,8\left(mol\right);n_{AgNO_3}=2\cdot0,1=0,2\left(mol\right)\\ PTHH:KCl+AgNO_3\rightarrow KNO_3+AgCl\downarrow\\ ....0,2....0,2.....0,2....0,2\left(mol\right)\)

Vì \(\dfrac{n_{KCl}}{1}>\dfrac{n_{AgNO_3}}{1}\) nên sau phản ứng KCl dư, tính theo AgNO₃

\(\Rightarrow m_{AgCl}=0,2\cdot143,5=28,7\left(g\right)\)

\(b,V_{dd\left(sau.p/ứ\right)}=V_{KNO_3}=V_{KCl}+V_{AgNO_3}=0,4+0,1=0,5\left(l\right)\\ \Rightarrow C_{M_{KNO_3}}=\dfrac{0,2}{0,5}=0,4M\)

1. \(CaCl_2+2AgNO_3\rightarrow Ca\left(NO_3\right)_2+2AgCl_{\downarrow}\)

\(n_{AgCl}=\dfrac{28,7}{143,5}=0,2\left(mol\right)\)

Theo PT: \(n_{Ca\left(NO_3\right)_2}=\dfrac{1}{2}n_{AgCl}=0,1\left(mol\right)\)

\(\Rightarrow C_{M_{Ca\left(NO_3\right)_2}}=\dfrac{0,1}{0,15+0,05}=0,5\left(M\right)\)

2. \(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)

BTNT Na, có: n_NaOH = n_Na = 0,1 (mol)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(2Al+2NaOH+2H_2O\rightarrow2NaAlO_2+3H_2\)

Xét tỉ lệ \(\dfrac{0,2}{2}>\dfrac{0,1}{2}\), ta được Al dư.

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{NaOH}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.22,4=3,36\left(l\right)\)

\(n_{Al\left(pư\right)}=n_{NaOH}=0,1\left(mol\right)\Rightarrow n_{Al\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)

\(\Rightarrow m_{Al\left(dư\right)}=0,1.27=2,7\left(g\right)\)

\(CuO+2HCl\rightarrow CuCl_2+H_2O\\ n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\\ n_{HCl}=2.0,1=0,2\left(mol\right)\\ V_{ddHCl}=\dfrac{0,2}{1}=0,2\left(l\right)\\ V_{ddsau}=V_{ddHCl}=0,2\left(l\right)\\ n_{CuCl_2}=n_{CuO}=0,2\left(mol\right)\\ C_{MddCuCl_2}=\dfrac{0,2}{0,2}=1\left(M\right)\)

Ta có: \(n_{Fe}=\dfrac{3,92}{56}=0,07\left(mol\right)\)

\(PTHH:Fe+CuSO_4--->FeSO_4+Cu\downarrow\)

a. Theo PT: \(n_{Cu}=n_{FeSO_4}=n_{Fe}=0,07\left(mol\right)\)

\(\Rightarrow m_{Cu}=0,07.64=4,48\left(g\right)\)

Ta có: \(V_{dd_{FeSO_4}}=V_{dd_{CuSO_4}}=\dfrac{200}{1000}=0,2\left(lít\right)\)

\(\Rightarrow C_{M_{FeSO_4}}=\dfrac{0,07}{0,2}=0,35M\)