PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
a, Giả sử: \(\left\{{}\begin{matrix}n_{Cu}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\)
⇒ 64x + 27y = 9,1 (1)
Ta có: \(n_{H_2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PT: \(n_{Al}=\frac{2}{3}n_{H_2}=0,1\left(mol\right)\)
⇒ y = 0,1 (2)
Từ (1) và (2) ⇒ x = y = 0,1 (mol)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Cu}=\frac{0,1.64}{9,1}.100\%\approx70,33\%\\\%m_{Al}\approx29,67\%\end{matrix}\right.\)
b,Theo PT: \(n_{H_2SO_4}=n_{H_2}=0,15\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,15.98=14,7\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\frac{14,7.100}{9,8}=150\left(g\right)\)
c, Ta có: m dd sau pư = mAl + m dd H2SO4 - mH2
= 0,1.27 + 150 - 0,15.2
= 152,4 (g)
Theo PT: \(n_{Al_2\left(SO_4\right)_3}=\frac{1}{2}n_{Al}=0,05\left(mol\right)\)
\(\Rightarrow C\%_{Al_2\left(SO_4\right)_3}=\frac{0,05.342}{152,4}.100\%\approx11,22\%\)
Bạn tham khảo nhé!
Cu không phản ứng với dd axit sunfuric loãng
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
a) Ta có: \(n_{H_2}=\frac{3,36}{22,4}=0,15\left(mol\right)\) \(\Rightarrow n_{Al}=0,1mol\)
\(\Rightarrow m_{Al}=27\cdot0,1=2,7\left(g\right)\) \(\Rightarrow m_{Cu}=6,4\left(g\right)\)
b) Theo PTHH: \(n_{H_2SO_4}=n_{H_2}=0,15mol\)
\(\Rightarrow m_{H_2SO_4}=0,15\cdot98=14,7\left(g\right)\) \(\Rightarrow m_{ddH_2SO_4}=\frac{14,7}{9,8\%}=150\left(g\right)\)
c) Ta có: \(n_{Al_2\left(SO_4\right)_3}=\frac{1}{2}n_{Al}=0,05mol\) \(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,05\cdot342=17,1\left(g\right)\)
Mặt khác: \(m_{H_2}=0,15\cdot2=0,3\left(g\right)\)
\(\Rightarrow m_{dd}=m_{hh}+m_{ddH_2SO_4}-m_{H_2}-m_{Cu}=152,4\left(g\right)\)
\(\Rightarrow C\%_{Al_2\left(SO_4\right)_3}=\frac{17,1}{152,4}\cdot100\approx11,22\)
