\(a,PTHH:\text{2Al + 3H2SO4 → Al2(SO4)3 + 3H2}\uparrow\)
\(\text{2Fe + 6H2SO4 → Fe2(SO4)3 + 6H2O + 3SO2}\uparrow\)
Cu không phản ứng.
Bạn tự viết tỉ lệ phương trình nhé :
\(b,nH2=0,4\left(mol\right)\)
Gọi số mol của Al , Fe lần lượt là x, y
\(\Rightarrow mCu=1\left(g\right)\)
\(\text{Nên : mAl+mFe=11}\)
\(\Rightarrow\left\{{}\begin{matrix}\text{27x+56y=11}\\\text{3/2x+y=0,4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(\Rightarrow\%Al=\frac{0,2.27.100}{11}=49,09\%\)
\(\Rightarrow\%Fe=\frac{0,1.\frac{3}{2}.100}{0,3}=50\%\)
\(c,\text{Theo PT nH2SO4=nH2=0,4}\)
\(\Rightarrow\text{mH2SO4=39,2}\)
\(\Rightarrow m_{dd}=\text{39,2.100/9,8=400 g}\)
Cu k phản ứng với H2SO4
\(\Rightarrow m_{Cu}=m_{cr}=1\left(g\right)\)
\(\Rightarrow m_{hhcl}=12-1=11\left(g\right)\)
\(n_{Al}=x;n_{Fe}=y\)
\(PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
Theo đề ta có:
\(hpt:\left\{{}\begin{matrix}27x+56y=11\\1,5x+y=\frac{8,96}{22,4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\frac{0,2.27}{11}.100\%=49,1\left(\%\right)\\\%m_{Fe}=100-49,1=50,9\left(\%\right)\end{matrix}\right.\)
\(m_{H_2SO_4}=98.\left(\frac{0,2.3}{2}+0,1\right)=39,2\left(g\right)\)
\(\rightarrow m_{dd}=\frac{39,2.100}{9,8}=400\left(g\right)\)
