Ta có: \(n_K=\dfrac{1,56}{39}=0,04\left(mol\right)\)

\(m_{CuSO_4}=200.8\%=16\left(g\right)\Rightarrow n_{CuSO_4}=\dfrac{16}{160}=0,1\left(mol\right)\)

PT: \(2K+2H_2O\rightarrow2KOH+H_2\)

___0,04___________0,04___0,02 (mol)

 \(2KOH+CuSO_4\rightarrow K_2SO_4+Cu\left(OH\right)_{2\downarrow}\)

Xét tỉ lệ: \(\dfrac{0,04}{2}< \dfrac{0,1}{1}\), ta được CuSO₄ dư.

Theo PT: \(n_{Cu\left(OH\right)_2}=n_{CuSO_4\left(pư\right)}=n_{K_2SO_4}=\dfrac{1}{2}n_{KOH}=0,02\left(mol\right)\)

⇒ n_{CuSO4 (dư)} = 0,08 (mol)

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

___0,02____0,02 (mol)

a, Ta có: V_B = 0,02.22,4 = 0,448 (l)

b, m_CuO = 0,02.80 = 1,6 (g)

c, Ta có: m _{dd sau pư} = 1,56 + 200 - 0,02.2 - 0,02.98 = 199,56 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{K_2SO_4}=\dfrac{0,02.174}{199,56}.100\%\approx1,74\%\\C\%_{CuSO_4\left(dư\right)}=\dfrac{0,08.160}{199,56}.100\%\approx6,41\%\end{matrix}\right.\)

Bạn tham khảo nhé!

\(n_K=\dfrac{1,56}{39}=0,04\left(mol\right)\)

\(n_{CuSO4}=\dfrac{200.8}{100.160}=0,1\left(mol\right)\)

\(2K+2H2O\rightarrow2KOH+H2\)

0,04--------------->0,04------------->0,02(mol)

\(2KOH+CuSO4\rightarrow Cu\left(OH\right)2+K2SO4\)

0,04-------->0,02----->0,02------->0,02(mol)

=> dd A gồm CuSO4 dư và K2SO4

=> Khí B là H2

=>Kết tủa C là Cu(OH)2

a) \(V_{H2}=0,02.22,4=0,448\left(l\right)\)

b)\(Cu\left(OH\right)2\rightarrow CuO+H2O\)

0,02---------------->0,02(mol)

\(m_{CuO}=0,02.80=1,6\left(g\right)\)

c) \(m_{KOH}=0,04.56=2,24\left(g\right)\)

\(m_{Cu\left(OH\right)2}=0,02.98=1,96\left(g\right)\)

\(\Rightarrow\)\(m_{dd}=200+2,24-1,96=200,28\left(g\right)\)

\(C\%_{K2SO4}=\dfrac{0,02.174}{200,28}.100\%=1,74\%\)

\(n_{CuSO4}dư=0,1-0,02=0,08\left(mol\right)\)

\(C\%_{CuSO4}=\dfrac{0,08.160}{200,28}.100\%=6,39\%\)

Chúc bạn học tốt^^

Pt: Ba+2H₂O -> Ba(OH)₂+H₂ (1)

Ba(OH)₂+CuSO₄ ->Cu(OH)₂ \(\downarrow\) +BaSO₄ \(\downarrow\)(2)

Ba(OH)₂+(NH₄)₂SO₄ ->BaSO₄ \(\downarrow\)+2NH₃+2H₂O (3)

Cu(OH)₂\(\underrightarrow{t^0}\)CuO+H₂O (4)

BaSO₄ \(\underrightarrow{t^0}\) ko xảy ra phản ứng

Theo (1) ta có \(n_{H_2}=n_{Ba\left(OH\right)_2}=n_{Ba}=\frac{27,4}{137}=0,2\left(mol\right)\)

\(n_{\left(NH_4\right)_2SO_4}=\frac{1,32\cdot500}{132\cdot100}=0,05\left(mol\right)\)

\(n_{CuSO_4}=\frac{2\cdot500}{100\cdot160}=0,0625\left(mol\right)\)

Ta thấy: \(n_{Ba\left(OH\right)_2}>n_{\left(NH_4\right)_2SO_4}+n_{CuSO4\:}\) nên Ba(OH)₂ dư và 2 muối đều phản ứng hết

Theo (2) ta có: \(n_{Ba\left(OH\right)_2}=n_{Cu\left(OH\right)_2}=n_{BaSO_4}=n_{CuSO_4}=0,0625\left(mol\right)\)

Theo (3) ta có: \(n_{Ba\left(OH\right)_2}=n_{BaSO_4}=n_{\left(NH_4\right)_2SO_4}=0,05\left(mol\right)\)

và \(n_{NH_3}=2n_{\left(NH_4\right)_2SO_4}=0,05\cdot2=0,1\left(mol\right)\)

\(\Rightarrow n_{Ba\left(OH_2\right)}\text{dư}=0,2-\left(0,05+0,0625\right)=0,0875\left(mol\right)\)

a)\(V_{A\left(ĐKTC\right)}=V_{H_2}+V_{NH_3}=\left(0,2+0,1\right)\cdot22,4=6,72\left(l\right)\)

b)Theo (4) ta có: \(n_{CuO}=n_{Cu\left(OH\right)_2}=0,0625\left(mol\right)\)

\(m_{\text{chất rắn}}=m_{BaSO_4}+m_{CuO}=\left(0,0625+0,05\right)\cdot233+0,0625\cdot80=31,2125\left(g\right)\)

\(n_{Na}=\dfrac{2.3}{23}=0.1\left(mol\right)\)

\(n_{HCl}=0.4\cdot2=0.8\left(mol\right)\)

\(NaOH+HCl\rightarrow NaCl+H_2O\)

\(0.8..............0.8\)

\(2Na+2H_2O\rightarrow2NaOH+H_{_{ }2}\)

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

\(n_{NaOH}=0.1< 0.8\)

Đề nhầm lãn !

Help meeee

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

a/

\(n_{Na_2O}=\dfrac{9,3}{62}=0,15\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

0,15 0,3 (mol)

\(m_{NaOH}=0,3.40=12\left(g\right)\)

\(m_A=90,7+9,3=100\left(g\right)\)

\(C\%_{NaOH}=\dfrac{12}{100}.100\%=12\%\)

b/

m\(_{FeSO_4}=\dfrac{16.200}{100}=32\left(g\right)\)

\(\rightarrow m_{FeSO_4}=\dfrac{32}{152}=\dfrac{4}{19}\left(mol\right)\)

\(2NaOH+FeSO_4\rightarrow Na_2SO_4+Fe\left(OH\right)_2\downarrow\)

bđ: 0,3 \(\dfrac{4}{19}\) 0 0 (mol)

pư: 0,3 0,15 0,15 0,15 (mol)

dư: 0 \(\dfrac{23}{380}\) (mol)

\(m_{Fe\left(OH\right)_2}=0,15.90=13,5\left(g\right)\)

\(m_C=100+200-13,5=286,5\left(g\right)\)

\(m_{Na_2SO_4}=0,15.142=21,3\left(g\right)\)

\(\rightarrow C\%_{Na_2SO_4}=\dfrac{21,3}{286,5}.100\%\approx7,4\%\)

\(m_{FeSO_4\left(dư\right)}=\dfrac{23}{380}.152=9,2\left(g\right)\)

\(\rightarrow C\%_{FeSO_4\left(dư\right)}=\dfrac{9,2}{286,5}.100\%\approx3,2\%\)

nNa = 6.9 : 23 = 0.3 mol

           4Na + O2 ->2 Na2O

mol :  0.3 ->           0.15

          Na2O + H2O -> 2NaOH

mol : 0.15 ->                0.3

mdd = 0.15 x 62 + 140.7 = 150g

C% NaOH = 0.3x40: 150 x 100% = 8%

gfvfvfvfvfvfvfv555