a) $2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$

b)

n H2SO4 = 0,03.1 = 0,03(mol)

n NaOH = 2n H2SO4 = 0,06(mol)

=> CM NaOH = 0,06/0,05 = 1,2M

c) $H_2SO_4 + 2KOH \to K_2SO_4 + 2H_2O$

n KOH = 2n H2SO4 = 0,06(mol)

=> m KOH = 0,06.56 = 3,36 gam

=> m dd KOH = 3,36/5,6% = 60(gam)

=> V dd KOH = m/D = 60/1,045 = 57,42(ml)

\(Đặt:m_{CuSO_4\left(thêm\right)}=a\left(g\right)\\ V_{ddCuSO_4\left(sau\right)}=\dfrac{a+160}{1,25}\\ C_{MddCuSO_4\left(sau\right)}=1\left(M\right)\\ \Leftrightarrow\dfrac{\dfrac{a+160.5\%}{160}}{\dfrac{a+160}{1,25.1000}}=1\\ \Leftrightarrow a=14,312\left(g\right)\\ \Rightarrow m_{CuSO_4\left(thêm\right)}=14,312\left(g\right)\)

a) C% NaCl = 10/(10 + 70)  .100% = 12,5%

b) n CuSO4 = 32/160 = 0,2(mol)

V H2O = 0,45(lít)

=> CM CuSO4 = 0,2/0,45 = 0,444M

C% NaCl = 10 / ( 10 + 70 ) * 100% = 12.5%

nCuSO4 = 32/160 = 0.2 (mol) 

CM CuSO4 = 0.2/0.45 = 0.44 (M)

\(m_{dd}=10+70=80\left(g\right)\)

=> C%=\(\dfrac{10}{80}.100\%=12,5\%\)

\(n_{CuSO_4}=\dfrac{32}{160}=0,2\left(mol\right)\)

đổi 450ml=0,45 lít

=>\(C_M=\dfrac{0,2}{0,45}=0,44\left(M\right)\)

Chúc bạn học tốt

nH2SO4=0,1(mol) 

a) PTHH: 2 NaOH + H2SO4 -> Na2SO4 + 2 H2O

b) 0,2___________0,1________0,1(mol)

mNaOH=0,2.40=8(g)

=>mddNaOH=(8.100)/25= 32(g)

c) mNa2SO4=0,1.142=14,2(g)

d) PTHH: 2 KOH + H2SO4 -> K2SO4 +2 H2O

nKOH=0,2(mol) => mKOH=0,2.56=11,2(g)

=> mddKOH=(11,2.100)/8=140(g)

=> VddKOH= 140/1,085=129,03(ml)

Chúc em học tốt!

a, \(2KOH+MgSO_4\rightarrow K_2SO_4+Mg\left(OH\right)_2\)

b, Ta có: \(n_{KOH}=0,2.1=0,2\left(mol\right)\)

Theo PT: \(n_{MgSO_4}=n_{K_2SO_4}=n_{Mg\left(OH\right)_2}=\dfrac{1}{2}n_{KOH}=0,1\left(mol\right)\)

\(\Rightarrow m_{Mg\left(OH\right)_2}=0,1.58=5,8\left(g\right)\)

c, \(V_{MgSO_4}=\dfrac{0,1}{2}=0,05\left(l\right)\)

d, \(C_{M_{K_2SO_4}}=\dfrac{0,1}{0,2+0,05}=0,4\left(M\right)\)

a, PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

b, Ta có: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,5.65=32,5\left(g\right)\)

\(\Rightarrow m_{CuO}=72,5-32,5=40\left(g\right)\)

c, Ta có: \(n_{CuO}=\dfrac{40}{80}=0,5\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{Zn}+n_{CuO}=1\left(mol\right)\)

\(\Rightarrow b=C_{M_{H_2SO_4}}=\dfrac{1}{2,5}=0,4M\)

c, Theo PT: \(\left\{{}\begin{matrix}n_{ZnSO_4}=n_{Zn}=0,5\left(mol\right)\\n_{CuSO_4}=n_{CuO}=0,5\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{ZnSO_4}}=\dfrac{0,5}{2,5}=0,2M\\C_{M_{CuSO_4}}=\dfrac{0,5}{2,5}=0,2M\end{matrix}\right.\)

Bạn tham khảo nhé!

a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\left(I\right)\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\left(II\right)\)

b, Theo PTHH(1) : \(n_{Zn}=n_{H_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{Zn}=32,5\left(g\right)\)

\(\Rightarrow m_{CuO}=m_{hh}-m_{Zn}=40\left(g\right)\)

\(\Rightarrow n_{CuO}=\dfrac{m}{M}=0,5\left(mol\right)\)

c, Theo PTHH (1) và (2) : \(n_{H2SO4}=n_{CuO}+n_{Zn}=1\left(mol\right)\)

\(\Rightarrow C_{MH2SO4}=b=\dfrac{n}{V}=\dfrac{1}{2,5}=0,4M\)

d, ( Chắc là thể tích coi như không đổi )

Thấy sau phản ứng thu được A gồm \(0,5molZnSO_4,0,5molCuSO_4\)

\(\Rightarrow C_{MCuSO4}=C_{MZnSO4}=\dfrac{n}{V}=\dfrac{0,5}{2,5}=0,2M\)

Vậy ...

\(n_{Zn}=\dfrac{4,55}{65}=0,07(mol)\\ Zn+2HCl\to ZnCl_2+H_2\\ a,n_{HCl}=0,14(mol)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,14}{0,2}=0,7M\\ b,n_{H_2}=0,07(mol)\\ \Rightarrow V_{H_2}=0,07.22,4=1,568(l)\\ c,n_{ZnCl_2}=0,07(mol)\\ \Rightarrow m_{ZnCl_2}=0,07.136=9,52(g)\\ c,ZnCl_2+2AgNO_3\to 2AgCl\downarrow+Zn(NO_3)_2\)

\(m_{dd_{ZnCl_2}}=200.0,8+4,55-0,07.2=164,41(g)\\ n_{AgCl}=0,14(mol);n_{Zn(NO_3)_2}=0,07(mol)\\ \Rightarrow C\%_{Zn(NO_3)_2}=\dfrac{0,07.189}{164,41+200-0,14.143,5}.100\%=3,84%\)

Câu a)

\(m_{ddCuSO_4\left(10\%\right)}=400.1,1=440\left(g\right)\\ m_{CuSO_4\left(dd.10\%\right)}=10\%.440=44\left(g\right)\\ C\%_{ddCuSO_4\left(cuối\right)}=20,8\%\\ \Leftrightarrow\dfrac{m_{CuSO_4}+44}{440+m_{CuSO_4}}.100\%=20,8\%\\ \Leftrightarrow m_{CuSO_4}=60\left(g\right)\)

Vậy: Cần lấy 60 gam CuSO₄ hoà tan vào 400 ml dung dịch CuSO₄ 10% (D=1,1g/ml) để tạo dung dịch C có nồng độ 20,8%

Câu b em xem link này he https://hoc24.vn/cau-hoi/acan-lay-bao-nhieu-g-cuso4-hoa-tan-vao-400ml-dd-cuso4-10d11gml-de-tao-thanh-dd-c-co-nong-do-288-b-khi-ha-nhiet-do-dd-c-xuong-12doc-thi-th.224557369474

\(a,2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ b,n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\\ \Rightarrow n_{Al}=0,2(mol)\\ \Rightarrow m_{Al}=0,2.27=5,4(g)\\ c,n_{H_2SO_4}=0,3(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,3.98}{9,8\%}=300(g)\\ d,n_{Al_2(SO_4)_3}=0,1(mol)\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{0,1.342}{5,4+300-0,3.2}.100\%=11,22\%\)