\(n_{KOH}=\dfrac{400.7}{100}:56=0,5\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{100.19,6}{100}:98=0,2\left(mol\right)\)

\(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

0,4             0,2            0,2

Lập tỉ lệ:

\(\dfrac{0,5}{2}>\dfrac{0,2}{1}\) => KOH dư.

\(m_{dd}=400+100=500\left(g\right)\)

\(n_{KOH.dư}=0,5-0,4=0,1\left(mol\right)\)

\(C\%_{K_2SO_4}=\dfrac{0,2.174.100}{500}=6,96\%\)

\(C\%_{KOH}=\dfrac{0,1.56.100}{500}=1,12\%\)

\(n_{H_2SO_4}=\dfrac{200.19,6}{100.98}=0,4mol\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ n_{CuSO_4\left(A\right)}=n_{CuO}=n_{H_2SO_4}=0,4mol\\ n_{Cu\left(OH\right)_2}=\dfrac{29,4}{98}=0,3mol\\ CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\\Rightarrow\dfrac{0,4}{1}>\dfrac{0,3}{1}\Rightarrow CuSO_4.pư.không.hết\)

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

0,3mol        0,6mol                                        0,3mol

\(m_{ddB}=0,4.80+200+0,6.40-29,4=226,6g\\ C_{\%Na_2SO_4\left(B\right)}=\dfrac{0,3.142}{226,6}\cdot100=18,8\%\)

\(\begin{cases} m_{H_2SO_4}=\dfrac{100.19,6\%}{100\%}=19,6(g)\\ m_{BaCl_2}=\dfrac{300.20,8\%}{100\%}=62,4(g) \end{cases} \Rightarrow \begin{cases} n_{H_2SO_4}=\dfrac{19,6}{98}=0,2(mol)\\ n_{BaCl_2}=\dfrac{62,4}{208}=0,3(mol) \end{cases}\\ a,PTHH:BaCl_2+H_2SO_4\to BaSO_4\downarrow +2HCl\)

Vì \(\dfrac{n_{H_2SO_4}}{1}<\dfrac{n_{BaCl_2}}{1}\) nên \(BaCl_2\) dư

\(\Rightarrow n_{BaSO_4}=0,2(mol)\\ \Rightarrow m_{BaSO_4}=0,2.233=46,6(g)\)

\(b,n_{HCl}=n_{BaSO_4}=0,2(mol)\\ \Rightarrow m_{CT_{HCl}}=0,2.36,5=7,3(g)\\ m_{dd_{HCl}}=100+300-46,6=353,4(g)\\ \Rightarrow C\%_{HCl}=\dfrac{7,3}{353,4}.100\%\approx 2,07\%\)

Ta có: \(n_{H_2SO_4}=\dfrac{200.19,6\%}{98}=0,4\left(mol\right)\)

\(n_{Cu\left(OH\right)_2}=\dfrac{29,4}{98}=0,3\left(mol\right)\)

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

Dung dịch A gồm: CuSO₄ và H₂SO₄ dư

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

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

Đề có cho dữ kiện gì liên quan đến dd NaOH không bạn nhỉ?

\(a,n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\\ m_{H_2SO_4}=200.19,6\%=39,2\left(g\right)\\ \rightarrow n_{H_2SO_4}=\dfrac{39,2}{98}=0,4\left(mol\right)\)

PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

bđ        0,3    0,4

pư        0,3        0,3

spư      0           0,1           0,3        0,3

\(\rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)

\(b,m_{dd}=19,5+200-0,3.2=218,9\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}C\%_{ZnSO_4}=\dfrac{0,3.161}{218,9}.100\%=22,06\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,1.98}{218,9}.100\%=4,48\%\end{matrix}\right.\)

Tham Khảo

Chọn đáp án A.

Đáp án D

Đáp án D