Question

Hòa tan 13,5%(g) Nhôm bằng 1 lượng vừa đủ dd H2SO4 24.5%.
a) Tính khối lượng dd axit đã dùng.
b )Tính thể tích khí thoát ra (ở đktc).
c) Tính nồng độ % của dd muối phản ứng.

Answer 1

2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

\(n_{Al}=\dfrac{13,5}{27}=0,5\left(mol\right)\)

a) Theo PT: \(n_{H_2SO_4}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}\times0,5=0,75\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,75\times98=73,5\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{73,5}{24,5\%}=300\left(g\right)\)

b) Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}\times0,5=0,75\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,75\times22,4=16,8\left(l\right)\)

\(m_{H_2}=0,75\times2=1,5\left(g\right)\)

c) Theo PT: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}\times0,5=0,25\left(mol\right)\)

\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,25\times342=85,5\left(g\right)\)

\(m_{ddAl_2\left(SO_4\right)_3}=13,5+300-1,5=312\left(g\right)\)

\(\Rightarrow C\%_{Al_2\left(SO_4\right)_3}=\dfrac{85,5}{312}\times100\%=27,4\%\)

Answer 2

2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

\(n_{Al}=\dfrac{13,5}{27}=0,5\left(mol\right)\)

theo PT: \(n_{H_2SO_4}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}\times0,5=0,75\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,75\times98=73,5\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{73,5}{24,5\%}=300\left(g\right)\)

Theo PT: \(n_{H_2}=n_{Al}=\dfrac{3}{2}\times0,5=0,75\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,75\times22,4=16,8\left(l\right)\)

\(m_{H_2}=0,75\times2=1,5\left(g\right)\)

\(m_{ddAl_2\left(SO_4\right)_3}=13,5+300-1,5=312\left(g\right)\)

Theo PT: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}\times0,5=0,25\left(mol\right)\)

\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,25\times342=85,5\left(g\right)\)

\(\Rightarrow C\%_{ZnSO_4}=\dfrac{85,5}{312}\times100\%=27,4\%\)