Question

cho 5,49(g) Al+200g dung dich H2So4

a) tinh v H2 sinh ra ở đktc

b)tính c%dung dịch sau p/ư

Answer 1

a) PTHH: 2Al + 3H₂SO₄ \(\rightarrow\) Al₂(SO₄)₃ + 3H₂\(\uparrow\)
n_Al = \(\frac{5,49}{27}=0,2\left(mol\right)\)
Theo PT: n_{\(H_2\)} = \(\frac{3}{2}\)n_Al = \(\frac{3}{2}.0,2=0,3\left(mol\right)\)
=> m_{\(H_2\)} = 0,3.2 = 0,6 (g)
=> V_{\(H_2\)} = 0,3.22,4 = 6,72 (l)
b) Theo PT: n_{\(Al_2\left(SO_4\right)_3\)} = \(\frac{1}{2}n_{Al}\) = \(\frac{1}{2}.0,2=0,1\left(mol\right)\)
=> m_{\(Al_2\left(SO_4\right)_3\)} = 0,1.342 = 34,2 (g)
=> m_{dd sau pứ} = 5,49 + 200 - 0,6 = 204,89 (g)
=> C%_{dd sau pứ} = \(\frac{34,2}{204,89}.100\%=16,7\%\)

Answer 2

Bạn sửa đề hộ mình là 5.4g Al

nAl= 5.4/27=0.2 mol

2Al + 3H2SO4 --> Al2(SO4)3 + 3H2

0.2_______________0.1________0.3

VH2= 0.3*22.4=6.72l

mdd sau phản ứng= mAl + mddH2SO4 - mH2= 5.4+200-0.3*2=204.8g

mAl2(SO4)3= 0.1*342=34.2g

C%Al2(SO4)3= 34.2/204.8*100%= 16.69%