CuO=>Cu(OH)2

Al2O3=>Al(OH)3BaO=>Ba(OH)2

nạo?

a)\(m_{H_2SO_4}=\frac{9,8.20}{100}=1,96\left(g\right)\)

\(n_{H_2SO_4}=\frac{m}{M}=\frac{1,96}{98}=0,02\left(mol\right)\)

\(m_{BaCl_2}=\frac{400.5,2}{100}=20,8\left(g\right)\)

\(n_{BaCl_2}=\frac{m}{M}=\frac{20,8}{208}=0,1\left(mol\right)\)

PTHH:

\(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)

bđ:0,1 0,02 0 0(mol)

pứ: 0,02 0,02 0,02 0,04(mol)

spứ: 0,08 0 0,02 0,04(mol)

\(m_{BaSO_4}=n.M=0,02.233=4,66\left(g\right)\)

b)\(m_{ddspu}=m_{ddH_2SO_4}+m_{ddBaCl_2}=9,8+400=409,8\left(g\right)\)

\(m_{BaCl_2}du=n.M=0,08.208=16,64\left(g\right)\)

\(m_{HCl}=n.M=0,04.36,5=1,46\left(g\right)\)

\(C\%_{BaCl_2}du=\frac{16,64.100}{409,8}=4,06\%\)

\(C\%_{HCl}=\frac{1,46.100}{409,8}=0,36\%\)

❤Lưu ý nhé: \(BaSO_4\) là kết tủa không tan trong dung dịch nên nếu tính nồng độ phần trăm là sai❤

a)\(SO_2;CaO;P_2O_5\)

PTHH:

\(SO_2+H_2O\rightarrow H_2SO_3\)

\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

b)\(CaO;Fe_3O_4\)

PTHH:

\(CaO+2HCl\rightarrow CaCl_2+H_2O\)

\(Fe_3O_4+8HCl\rightarrow FeCl_2+2FeCl_3+4H_2O\)

c)\(SO_2;SiO_2;P_2O_5\)

PTHH:

\(SO_2+2KOH\rightarrow K_2SO_3+H_2O\)

\(SO_2+KOH\rightarrow KHSO_3\)

\(P_2O_5+H_2O+2KOH\rightarrow2KH_2PO_4\)

\(P_2O_5+4KOH\rightarrow2K_2HPO_4+H_2O\)

\(P_2O_5+6KOH\rightarrow2K_3PO_4+3H_2O\)

\(2KOH+SiO_2\rightarrow K_2SiO_3+H_2O\)

a)Goi \(n_{CaCO_3}=x\left(mol\right);n_{K_2CO_3}=y\left(mol\right)\)

\(\Rightarrow100x+138y=6,86\left(I\right)\)

PTHH:

\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2\uparrow+H_2O\left(1\right)\)

x 2x x x x(mol)

\(K_2CO_3+2HCl\rightarrow2KCl+CO_2+H_2O\left(2\right)\)

y 2y 2y y y(mol)

\(n_{CO_2}=\frac{m}{M}=\frac{6,24}{44}=0,14\left(mol\right)\)

\(\Rightarrow x+y=0,14\left(mol\right)\left(II\right)\)

Tu (I), (II)\(\Rightarrow\left\{{}\begin{matrix}100x+138y=6,76\\x+y=0,14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,33\\y=-0,19\left(?\right)\end{matrix}\right.\)

De sai

Cau 4 :

Goi \(n_{NaCl}=x\left(mol\right);n_{Na_2SO_4}=y\left(mol\right)\)

PTHH:

\(2NaCl+Ba\left(NO_3\right)_2\rightarrow BaCl_2+2NaNO_3\left(1\right)\)

\(x;\frac{x}{2};\frac{x}{2};x\left(mol\right)\)

\(Na_2SO_4+Ba\left(NO_3\right)_2\rightarrow2NaNO_3+BaSO_4\downarrow\left(2\right)\)

\(y;y;2y;y\left(mol\right)\)

n\(n_{BaSO_4}=\frac{m}{M}=\frac{49,25}{233}0,21\left(mol\right)\)

\(\Rightarrow y=0,21\left(mol\right)\)

\(m_{Na_2SO_4}=n.M=0,21.142=29,82\left(g\right)\)

\(\%m_{Na_2SO_4}=\frac{29,82}{85}.10=35,08\%\)

\(\Rightarrow\%m_{NaCl}=100\%-35,08\%=64,92\%\)

Cau 3:

a) \(K_2O+H_2O\rightarrow2KOH\)

\(BaO+2H_2O\rightarrow Ba\left(OH\right)_2\)

b)\(2KOH+MgCl_2\rightarrow Mg\left(OH\right)_2+2KCl\)

\(3KOH+FeCl_3\rightarrow Fe\left(OH\right)_3+3KCl\)

\(Ba\left(OH\right)_2+MgCl_2\rightarrow Mg\left(OH\right)_2\downarrow+BaCl_2\)

\(3Ba\left(OH\right)_2+2FeCl_3\rightarrow2Fe\left(OH\right)_3\downarrow+3BaCl_2\)

Cau 2:

a)\(FeCl_2+2NaOH\rightarrow2NaCl+Fe\left(OH\right)_2\)

b)\(Ba\left(NO_3\right)_2+H_2SO_4\rightarrow BaSO_4\downarrow+HNO_3\)

c)ko pu
d)\(AgNO_3+NaCl\rightarrow AgCl\downarrow+NaNO_3\)

g)ko pu

e)\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

Cau 1

a)\(2NaCl+2H_2O\rightarrow^{dp}_{cmn}2NaOH+H_2+Cl_2\left(1\right)\)

\(2NaOH+MgSO_4\rightarrow Mg\left(OH\right)_2+Na_2SO_4\left(2\right)\)

\(Mg\left(OH\right)_2\rightarrow^{t^o}MgO+H_2O\left(3\right)\)

\(MgO+2HCl\rightarrow MgCl_2+H_2O\left(4\right)\)

b)\(N_2O\)? Sua lai: \(N_2O\rightarrow Na_2O\)

\(Na_2O+H_2O\rightarrow2NaOH\left(1\right)\)

\(NaOH+HCl\rightarrow NaCl+H_2O\left(2\right)\)

\(2H_2O+2NaCl\rightarrow^{dp}_{cmn}2NaOH+H_2+Cl_2\left(3\right)\)

\(NaOH+CO_2\rightarrow NaHCO_3\left(4\right)\)

a)\(m_{AgNO3}bd=\frac{250.4}{100}=10g\)

\(m_{AgNO3}pu=\frac{10.17}{100}=1,7g\)

\(n_{AgNO3}pu=\frac{m}{M}=\frac{1,7}{170}=0,01\left(mol\right)\)

PTHH:

\(Cu+2AgNO_3\rightarrow Cu\left(NO_3\right)_2+2Ag\downarrow\)

0,005 0,01 0,005 0,01 (mol)

\(m_{thanhCu}spu=m_{thanhCu}bd+m_{Ag\downarrow}-m_{Cu}pu=5+0,01.108+0,005.64=5,76g\)