Question

Cho 5,4 g bột nhôm vào 180g dung dịch H₂SO₄ 24,5%.

a) Tính thể tích khí H₂ sinh ra sau khi phản ứng kết thúc.

b) Tính C% của dung dịch

Answer 1

2Al + 3H₂SO₄ \(\rightarrow\)Al₂(SO₄)₃ + 3H₂

n_Al=\(\dfrac{5,4}{27}=0,2\left(mol\right)\)

m_H2SO4=\(180.\dfrac{24,5}{100}=44,1\left(g\right)\)

n_H2SO4=\(\dfrac{44,1}{98}=0,45\left(mol\right)\)

Vì 0,3<0,45 nên H₂SO₄ dư 0,15mol

Theo PTHH ta có:

\(\dfrac{3}{2}\)n_Al=n_H2=0,3(mol)

V_H2=22,4.0,3=6,72(lít)

b;

Theo PTHH ta có:

\(\dfrac{1}{2}\)n_Al=n_Al2(SO4)3=0,1(mol)

m_Al2(SO4)3=342.0,1=34,2(g)

m_H2SO4=0,15.98=14,7(g)

C% dd Al2(SO4)3=\(\dfrac{34,2}{5,4+180-0,3.2}.100\%=18,5\%\)

C% dd H₂SO₄=\(\dfrac{14,7}{5,4+180-0,6}.100\%=7,95\%\)

Answer 2

a) PTHH: 2Al + 3H₂SO₄ ----> Al₂(SO₄)₃ + 3H₂\(\uparrow\)

n_\(Al\) = \(\dfrac{5,4}{27}=0,2\left(mol\right)\)

m_{\(H_2SO_4\)} = \(\dfrac{24,5\%.180}{100\%}=44,1\left(g\right)\)

=> n_{\(H_2SO_4\)} = \(\dfrac{44,1}{98}=0,45\left(mol\right)\)

Ta có tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,45}{3}\) => Al phản ứng hết, H2SO4 dư

Theo PTHH: n_{\(H_2\)} = \(\dfrac{3}{2}n_{Al}\) = \(\dfrac{3}{2}.0,2=0,3\left(mol\right)\)

=> V_{\(H_2\)} = 0,3.22,4 = 6,72 (lít)

b) m_\(ddspu\) = 5,4 + 180 - 0,3.2 = 184,8 (g)

Theo PTHH: n_{\(Al_2\left(SO_4\right)_3\)} = \(\dfrac{1}{2}n_{Al}\) = \(\dfrac{1}{2}.0,2=0,1\left(mol\right)\)

=> m_{\(Al_2\left(SO_4\right)_3\)} = 0,1.342 = 34,2 (g)

C%_{\(Al_2\left(SO_4\right)_3\)} = \(\dfrac{34,2}{184,8}.100\%\approx18,5\%\)

Theo PTHH: n_H2SO4(p/ứ) = \(\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\)

=> m_H2SO4(dư) = ( 0,45 - 0,3 ). 98 = 14,7 (g)

=> C%_H2SO4(dư) = \(\dfrac{14,7}{184,8}.100\%\approx7,95\%\)