a)

Khối lượng của dung dịch:

\(m_{dd}=m_{ct}+m_{dm}=20+180=200\left(g\right)\)

Nồng độ phần trăm của dung dịch:

\(C\%=\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{20}{200}.100\%=10\%\)

b) đề sai nha bạn

Đặt : \(n_{CuO}=a\left(mol\right),n_{ZnO}=b\left(mol\right)\)

\(\Rightarrow80a+81b=28,25g\left(1\right)\)

a) Pt : \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

b) Ta có : \(n_{HCl}=\dfrac{20\%.127,75}{100\%.36,5}=0,7\left(mol\right)\Rightarrow2a+2b=0.7\left(2\right)\)

Từ (1),(2) \(\Rightarrow\left\{{}\begin{matrix}a=0,1=n_{CuCl2}\\b=0,25=n_{ZnCl2}\end{matrix}\right.\)

c) \(m_{muối}=m_{CuCl2}+m_{ZnCl2}=0,1.135+0,25.136=47,5\left(g\right)\)

d) \(\left\{{}\begin{matrix}C\%_{CuCl2}=\dfrac{0,1.135}{28,25+127,75}.100\%=8,65\%\\C\%_{ZnCl2}=\dfrac{0,25.136}{28,25+127,75}.100\%=21,79\%\end{matrix}\right.\)

\(m_{ct}=\dfrac{10.200}{100}=20\left(g\right)\)

\(n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)

a) Pt : \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O|\)

                  2              1                1               2

                 0,5           0,25          0,25

\(n_{H2SO4}=\dfrac{0,5.1}{2}=0,25\left(mol\right)\)

\(m_{H2SO4}=0,25.98=24,5\left(g\right)\)

\(m_{ddH2SO4}=\dfrac{24,5.100}{20}=122,5\left(g\right)\)

b) \(n_{Na2SO4}=\dfrac{0,25.1}{1}=0,25\left(mol\right)\)

⇒ \(m_{Na2SO4}=0,25.142=35,5\left(g\right)\)

\(m_{ddspu}=200+122,5=322,5\left(g\right)\)

\(C_{Na2SO4}=\dfrac{35,5.100}{322,5}=11\)⁰/₀

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\(n_{CuO}=\dfrac{1.6}{80}=0.02\left(mol\right)\)

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

\(0.02.......0.02.................0.02\)

\(m_{H_2SO_4}=0.02\cdot98=1.96\left(g\right)\)

\(m_{dd_{H_2SO_4}}=\dfrac{1.96}{20\%}=9.8\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng }}=1.6+9.8=11.4\left(g\right)\)

\(C\%_{CuSO_4}=\dfrac{0.02\cdot160}{11.4}=28.07\%\)

a) nCuSO4.5H2O=0,15(mol)

-> nCuSO4=0,15(mol) -> mCuSO4=160.0,15= 24(g)

mddCuSO4(sau)= 37,5+ 162,5=200(g)

C%ddCuSO4(sau)= (24/200).100= 12%

b) mCuSO4(tách)= (200/100) x 10=20(g)

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

                  a_____2a_______a_______a     (mol)

           \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

                  b_____6b_______2b_______3b   (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}40a+102b=18,2\\2a+6b=\dfrac{182,5\cdot20\%}{36,5}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgO}=\dfrac{0,2\cdot40}{18,2}\cdot100\%\approx43,96\%\\\%m_{Al_2O_3}=56,04\%\end{matrix}\right.\)

Theo PTHH: \(n_{MgCl_2}=0,2\left(mol\right)=n_{AlCl_3}\) \(\Rightarrow\left\{{}\begin{matrix}m_{MgCl_2}=0,2\cdot95=19\left(g\right)\\m_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{19}{18,2+182,5}\cdot100\%\approx9,47\%\\C\%_{AlCl_3}=\dfrac{26,7}{182,5+18,2}\cdot100\%\approx13,3\%\end{matrix}\right.\)

a) mHCl=182,5. 20%=36,5(g) -> nHCl=1(mol)

Đặt nMgO=a(mol); nAl2O3=b(mol)

PTHH: MgO +2 HCl -> MgCl2 + H2O

a__________2a______a(mol)

Al2O3 + 6 HCl ->  2 AlCl3 + 3 H2O

b_______6b______2b(mol)

b) Ta có hpt:

\(\left\{{}\begin{matrix}40a+102b=18,2\\2a+6b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)

=> mMgO=0,2.40=8(g)

=>%mMgO=(8/18,2).100=43,956%

=> %mAl2O3= 56,044%

c) m(muối)= mAlCl3 + mMgCl2= 133,5.2b+ 95.a= 133,5.0,1.2+95.0,2= 45,7(g)

d) mAlCl3= 26,7(g) ; mMgCl2 = 19(g)

mddsau= 18,2+ 182,5= 200,7(g)

=>C%ddAlCl3=(26,7/200,7).100=13,303%

C%ddMgCl2=(19/200,7).100=9,467%

giup mk voi

a, \(D=\dfrac{m}{V}\Rightarrow m_{ddNa_2SO_4}=1,12.2.1000=2240\left(g\right)\)

b, PT: \(BaCl_2+Na_2SO_4\rightarrow2NaCl+BaSO_{4\downarrow}\)

Ta có: \(m_{BaCl_2}=416.20\%=83,2\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{83,2}{208}=0,4\left(mol\right)\)

Theo PT: \(n_{BaSO_4}=n_{BaCl_2}=0,4\left(mol\right)\)

\(n_{NaCl}=2n_{BaCl_2}=0,8\left(mol\right)\)

Có: m _{dd sau pư} = 416 + 2240 - 0,4.233 = 2562,8 (g)

\(\Rightarrow C\%_{NaCl}=\dfrac{0,8.58,5}{2562,8}.100\%\approx1,83\%\)

\(n_{BaCl2}=\dfrac{20\%.416}{100\%.208}=0,4\left(mol\right)\)

 Pt : \(BaCl_2+Na_2SO_4\rightarrow BaSO_4\downarrow+2NaCl\)

            0,4                              0,4           0,8

a) \(m_{ddNa2SO4}=D.V=1,12.2000=2240\left(g\right)\)

b) \(m_{ddspu}=416+2240-\left(0,4.233\right)=2562,8\left(g\right)\)

\(C\%=\dfrac{\left(0,8.58,5\right)}{2562,8}.100\%=1,83\%\)

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