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Question

Bài 1: hòa tan 30,6 gam BaO trong 300ml dung dịch H2S04 2M . hãy tính khối lượng kết tủa thu được sau phản ứng ?

Bài 2:hòa tan hoàn toàn 16 gam SO3 vào nước được 200ml dung dịch A

a)tính CM của dung...

𝓓𝓾𝔂 𝓐𝓷𝓱 · Accepted Answer

Bài 1:

PTHH: $BaO+H_2SO_4\rightarrow BaSO_4\downarrow+H_2O$

Ta có: $\left\{{}\begin{matrix}n_{BaO}=\frac{30,6}{153}=0,2\left(mol\right)\n_{H_2SO_4}=0,3\cdot2=0,6\left(mol\right)\end{matrix}\right.$

Xét tỷ lệ: $\frac{0,2}{1}< \frac{0,6}{1}$ $\Rightarrow$ BaO phản ứng hết

$\Rightarrow n_{BaSO_4}=0,2mol$ $\Rightarrow m_{BaSO_4}=0,2\cdot233=46,6\left(g\right)$Câu trả lời: Bài 1:

PTHH: $BaO+H_2SO_4\rightarrow BaSO_4\downarrow+H_2O$

Ta có: $\left\{{}\begin{matrix}n_{BaO}=\frac{30,6}{153}=0,2\left(mol\right)\n_{H_2SO_4}=0,3\cdot2=0,6\left(mol\right)\end{matrix}\right.$

Xét tỷ lệ: $\frac{0,2}{1}< \frac{0,6}{1}$ $\Rightarrow$ BaO phản ứng hết

$\Rightarrow n_{BaSO_4}=0,2mol$ $\Rightarrow m_{BaSO_4}=0,2\cdot233=46,6\left(g\right)$

𝓓𝓾𝔂 𝓐𝓷𝓱 · Answer

Bài 2:

PTHH:  $SO_3+H_2O\rightarrow H_2SO_4$

a) Ta có: $n_{SO_3}=\frac{16}{80}=0,2\left(mol\right)$

$\Rightarrow n_{H_2SO_4}=0,2mol$ $\Rightarrow C_{M_{H_2SO_4}}=\frac{0,2}{0,2}=1\left(M\right)$

b) PTHH: $CuO+H_2SO_4\rightarrow CuSO_4+H_2O$

Ta có: $n_{CuO}=\frac{12}{80}=0,15\left(mol\right)$ $\Rightarrow$ CuO phản ứng hết

$\Rightarrow\left\{{}\begin{matrix}n_{CuSO_4}=0,15mol\n_{H_2SO_4\left(dư\right)}=0,05mol\end{matrix}\right.$ $\Rightarrow\left\{{}\begin{matrix}m_{CuSO_4}=0,15\cdot160=24\left(g\right)\m_{H_2SO_4\left(dư\right)}=0,05\cdot98=4,9\left(g\right)\end{matrix}\right.$

Mặt khác: $D_{nước}=1g/ml$  $\Rightarrow m_{H_2O}=200\left(g\right)$

$\Rightarrow m_{dd}=m_{SO_3}+m_{H_2O}+m_{CuO}=16+200+12=228\left(g\right)$

$\Rightarrow\left\{{}\begin{matrix}C\%_{CuSO_4}=\frac{24}{228}\cdot100\approx10,53\%\C\%_{H_2SO_4\left(dư\right)}=\frac{4,9}{228}\cdot100\approx2,15\%\end{matrix}\right.$Câu trả lời: Bài 2:

PTHH:  $SO_3+H_2O\rightarrow H_2SO_4$

a) Ta có: $n_{SO_3}=\frac{16}{80}=0,2\left(mol\right)$

$\Rightarrow n_{H_2SO_4}=0,2mol$ $\Rightarrow C_{M_{H_2SO_4}}=\frac{0,2}{0,2}=1\left(M\right)$

b) PTHH: $CuO+H_2SO_4\rightarrow CuSO_4+H_2O$

Ta có: $n_{CuO}=\frac{12}{80}=0,15\left(mol\right)$ $\Rightarrow$ CuO phản ứng hết

$\Rightarrow\left\{{}\begin{matrix}n_{CuSO_4}=0,15mol\n_{H_2SO_4\left(dư\right)}=0,05mol\end{matrix}\right.$ $\Rightarrow\left\{{}\begin{matrix}m_{CuSO_4}=0,15\cdot160=24\left(g\right)\m_{H_2SO_4\left(dư\right)}=0,05\cdot98=4,9\left(g\right)\end{matrix}\right.$

Mặt khác: $D_{nước}=1g/ml$  $\Rightarrow m_{H_2O}=200\left(g\right)$

$\Rightarrow m_{dd}=m_{SO_3}+m_{H_2O}+m_{CuO}=16+200+12=228\left(g\right)$

$\Rightarrow\left\{{}\begin{matrix}C\%_{CuSO_4}=\frac{24}{228}\cdot100\approx10,53\%\C\%_{H_2SO_4\left(dư\right)}=\frac{4,9}{228}\cdot100\approx2,15\%\end{matrix}\right.$

𝓓𝓾𝔂 𝓐𝓷𝓱 · Answer

Bài 3:

PTHH: $Mg+2HCl\rightarrow MgCl_2+H_2\uparrow$  (1)

$CuO+2HCl\rightarrow CuCl_2+H_2O$  (2)

a) Ta có: $n_{H_2}=\frac{4,48}{22,4}=0,2\left(mol\right)$

$\Rightarrow n_{Mg}=0,2mol$ $\Rightarrow m_{Mg}=0,2\cdot24=4,8\left(g\right)$

$\Rightarrow\%m_{Mg}=\frac{4,8}{12,8}\cdot100=37,5\%$ $\Rightarrow\%m_{CuO}=62,5\%$

b) Ta có: $\left\{{}\begin{matrix}n_{HCl\left(1\right)}=2n_{Mg}=0,4mol\n_{HCl\left(2\right)}=2n_{CuO}=2\cdot\frac{12,8-4,8}{80}=0,2mol\end{matrix}\right.$

$\Rightarrow n_{HCl}=0,6mol$ $\Rightarrow m_{HCl}=0,6\cdot36,5=21,9\left(g\right)$

Mặt khác: $m_{H_2}=0,2\cdot2=0,4\left(g\right)$

$\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{H_2}=12,8+200-0,4=212,4\left(g\right)$

$\Rightarrow C\%_{HCl}=\frac{21,9}{212,4}\cdot100\approx10,31\%$Câu trả lời: Bài 3: