Question

[NGUYỄN HƯƠNG GIANG]

cho mik hỏi

cho 25g hỗn hợp ( cu, al) td hết vs dd hcl -----> 0,56l khí đktc

a) pthh

b) mCu, mAl trong hỗn hợp

c) % mCu, %mAl

d) nếu cho lượng Cu, Al trên nung nóng vs 200g dd h2so4 đặc thì sẽ thu được bao nhiêu lít khí so2 ở đktc

Answer 1

[Minh Nhân]

\(n_{H_2}=\dfrac{0.56}{22.4}=0.025\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(n_{Al}=\dfrac{2}{3}\cdot0.025=\dfrac{1}{60}\left(mol\right)\)

\(m_{Al}=\dfrac{1}{60}\cdot27=0.45\left(g\right)\)

\(m_{Cu}=25-0.45=24.55\left(g\right)\)

\(\%Cu=\dfrac{24.55}{25}\cdot100\%=98.2\%\)

\(\%Al=100-98.2=1.8\%\)

\(Cu+2H_2SO_{4\left(đ\right)}\rightarrow CuSO_4+SO_2+2H_2O\)

\(2Al+6H_2SO_{4\left(đ\right)}\rightarrow Al_2\left(SO_4\right)_3+3SO_2+6H_2O\)

\(n_{Cu}=\dfrac{24.55}{64}=\dfrac{491}{1280}\left(mol\right)\)

\(V_{SO_2}=\left(\dfrac{1}{60}\cdot\dfrac{3}{2}+\dfrac{491}{1280}\right)\cdot22.4=9.1525\left(l\right)\)

Answer 2

[hnamyuh]

a)

$2Al + 6HCl \to 2AlCl_3 + 3H_2$
b)

$n_{H_2} = \dfrac{0,56}{22,4} = 0,225(mol)$

Theo PTHH : $n_{Al} = \dfrac{2}{3}n_{H_2} = \dfrac{1}{60}(mol)$
$m_{Al} = \dfrac{1}{60}.27 = 0,45(gam)$
$m_{Cu} = 25 - 0,45 = 24,55(gam)$

c)

$\%m_{Al} = \dfrac{0,45}{25}.100\% = 1,8\%$
$\%m_{Cu} = 100\% -1,8\% = 98,2\%$

d)

$Cu + 2H_2SO_4 \to CuSO_4 + SO_2 + 2H_2O$
$2Al + 6H_2SO_4 \to Al_2(SO_4)_3 + 3SO_2 + 6H_2O$

Theo PTHH : 

$n_{SO_2} = n_{Cu} + \dfrac{3}{2}n_{Al} = \dfrac{24,55}{64} + \dfrac{1}{60}.\dfrac{3}{2} = 0,41(mol)$
$V_{SO_2} = 0,41.22,4 = 9,184(lít)$