Question

[Thủy Nguyễn]

Cho 5,5g hh fe và al tác dụng dd hcl 14,6%-> 4,48l h2

A: tp%

B: mdd hcl

C: c% mỗi muối trong dd

Answer 1

[Minh Nhân]

Đặt :

nFe = x mol

nAl = y mol

Fe + 2HCl --> FeCl2 + H2

x____________x______x

2Al + 6HCl --> 2AlCl3 + 3H2

y______________y______1.5y

mhh= 56x + 27y = 5.5 g (1)

nH2 = x + 1.5y = 0.2 (2)

Giải (1) và (2) :

x = 0.05

y= 0.1

mFe = 2.8 g

mAl = 2.7 g

%Fe = 50.91%

%Al = 49.09%

Từ PTHH ta thấy :

nHCl = 2nH2 = 0.2*2=0.4 mol

mHCl = 0.4*36.5 = 14.6 g

mddHCl = 14.6*100/14.6 = 100g

mdd sau phản ứng = 5.5 + 100 - 0.4 = 105.1 g

mFeCl2 = 6.35g

mAlCl3 = 13.35g

C%FeCl2 = 6.04%

%AlCl3 = 12.7%

Answer 2

[Phùng Hà Châu]

Fe + 2HCl → FeCl₂ + H₂ (1)

2Al + 6HCl → 2AlCl₃ + 3H₂ (2)

\(n_{H_2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)

a) Gọi x,y lần lượt là số mol của Fe và Al

Ta có: \(56x+27y=5,5\) (*)

Theo Pt1: \(n_{H_2}=n_{Fe}=x\left(mol\right)\)

Theo PT2: \(n_{H_2}=\frac{3}{2}n_{Al}=1,5y\left(mol\right)\)

Ta có: \(x+1,5y=0,2\) (**)

Từ (*)(**) ta có: \(\left\{{}\begin{matrix}56x+27y=5,5\\x+1,5y=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)

Vậy \(n_{Fe}=0,05\left(mol\right)\Rightarrow m_{Fe}=0,05\times56=2,8\left(g\right)\)

\(n_{Al}=0,1\left(mol\right)\Rightarrow m_{Al}=0,1\times27=2,7\left(g\right)\)

\(\%m_{Fe}=\frac{2,8}{5,5}\times100\%=50,91\%\)

\(\%m_{Al}=100\%-50,91\%=49,09\%\)

b) Theo Pt1: \(n_{HCl}=2n_{Fe}=2\times0,05=0,1\left(mol\right)\)

Theo Pt2: \(n_{HCl}=3n_{Al}=3\times0,1=0,3\left(mol\right)\)

\(\Rightarrow\Sigma n_{HCl}=0,1+0,3=0,4\left(mol\right)\)

\(\Rightarrow\Sigma m_{HCl}=0,4\times36,5=14,6\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\frac{14,6}{14,6\%}=100\left(g\right)\)

c) \(m_{H_2}=0,2\times2=0,4\left(g\right)\)

Ta có: \(m_{dd}saupứ=5,5+100-0,4=105,1\left(g\right)\)

Theo Pt1: \(n_{FeCl_2}=n_{Fe}=0,05\left(mol\right)\)

\(\Rightarrow m_{FeCl_2}=0,05\times127=6,35\left(g\right)\)

\(\Rightarrow C\%_{FeCl_2}=\frac{6,35}{105,1}\times100\%=6,04\%\)

Theo Pt2: \(n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=0,1\times133,5=13,35\left(g\right)\)

\(\Rightarrow C\%_{AlCl_3}=\frac{13,35}{105,1}\times100\%=12,7\%\)