PT: \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
a, Ta có: \(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)
\(m_{H_2SO_4}=\dfrac{200.20}{100}=40\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{40}{98}=\dfrac{20}{49}\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{\dfrac{20}{49}}{3}\), ta được H2SO4 dư.
Theo PT: \(n_{H_2SO_4\left(pư\right)}=3n_{Al_2O_3}=0,3\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4\left(dư\right)}=\dfrac{53}{490}\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4\left(dư\right)}=\dfrac{53}{490}.98=10,6\left(g\right)\)
b, Theo PT: \(n_{Al_2\left(SO_4\right)_3}=n_{Al_2O_3}=0,1\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\)
c, Ta có: m dd sau pư = mAl2O3 + m dd H2SO4 = 210,2 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{H_2SO_4\left(dư\right)}=\dfrac{10,6}{210,2}.100\%\approx5,04\%\\C\%_{Al_2\left(SO_4\right)_3}=\dfrac{34,2}{210,2}.100\%\approx16,3\%\end{matrix}\right.\)
Bạn tham khảo nhé!
PT: A l 2 O 3 + 3 H 2 S O 4 → A l 2 ( S O 4 ) 3 + 3 H 2 O Al 2 O 3 +3H 2 SO 4 →Al 2 (SO 4 ) 3 +3H 2 O a, Ta có: n A l 2 O 3 = 10 , 2 102 = 0 , 1 ( m o l ) n Al 2 O 3 = 102 10,2 =0,1(mol) m H 2 S O 4 = 200.20 100 = 40 ( g ) ⇒ n H 2 S O 4 = 40 98 = 20 49 ( m o l ) m H 2 SO 4 = 100 200.20 =40(g)⇒n H 2 SO 4 = 98 40 = 49 20 (mol) Xét tỉ lệ: 0 , 1 1 < 20 49 3 1 0,1 < 3 49 20 , ta được H2SO4 dư. Theo PT: n H 2 S O 4 ( p ư ) = 3 n A l 2 O 3 = 0 , 3 ( m o l ) n H 2 SO 4 (pư) =3n Al 2 O 3 =0,3(mol) ⇒ n H 2 S O 4 ( d ư ) = 53 490 ( m o l ) ⇒n H 2 SO 4 (dư) = 490 53 (mol) ⇒ m H 2 S O 4 ( d ư ) = 53 490 . 98 = 10 , 6 ( g ) ⇒m H 2 SO 4 (dư) = 490 53 .98=10,6(g) b, Theo PT: n A l 2 ( S O 4 ) 3 = n A l 2 O 3 = 0 , 1 ( m o l ) n Al 2 (SO 4 ) 3 =n Al 2 O 3 =0,1(mol) ⇒ m A l 2 ( S O 4 ) 3 = 0 , 1.342 = 34 , 2 ( g ) ⇒m Al 2 (SO 4 ) 3 =0,1.342=34,2(g) c, Ta có: m dd sau pư = mAl2O3 + m dd H2SO4 = 210,2 (g) ⇒ { C % H 2 S O 4 ( d ư ) = 10 , 6 210 , 2 . 100 % ≈ 5 , 04 % C % A l 2 ( S O 4 ) 3 = 34 , 2 210 , 2 . 100 % ≈ 16 , 3 % ⇒ ⎩ ⎨ ⎧ C% H 2 SO 4 (dư) = 210,2 10,6 .100%≈5,04% C% Al 2 (SO 4 ) 3 = 210,2 34,2 .100%≈16,3%
