Question

Cho 100 g dd XCl3 16,25% pư hoàn toàn với 150g dd NaOH 8%

a, Xác định muối XCl3

b, Tính nồng độ dd sau pư

Answer 1

XCl3 + 3NaOH --> X(OH)3 + 3NaCl (1)

a) nXCl3=\(\dfrac{16,25}{MX+106,5}\left(mol\right)\)

nNaOH=0,3(mol)

theo (1) : nXCl3=1/3nNaOH=0,1(mol)

=>\(\dfrac{16,25}{MX+106,5}=0,1\)

=>M_X=56(g/mol)

=>X:Fe => XCl3 : FeCl3

b) nFeCl3=0,1(mol)

theo (1) : nFe(OH)3=nFeCl3=0,1(mol)

=>mFe(OH)3=10,7(g)

mdd sau pư=100+150-10,7=239,3(g)
theo (1) : nNaCl=3nFeCl3=0,3(mol)

=>mNaCl=17,55(g)

=>C%dd NaCl=7,4(%)

Answer 2

m_{XCl\(_3\)} = 100 . 16,25 % = 16,25 ( g )

m_NaOH = 150 . 8 % = 12 ( g )

\(\Rightarrow\) n_NaOH = \(\dfrac{12}{40}\) = 0,3 (mol)

PTHH : XCl₃ + 3NaOH \(\rightarrow\) 3NaCl + X(OH)₃

Theo PT có : 1mol XCl₃ : 3mol NaOH

\(\Rightarrow\) 0,3 mol NaOH : 0,1 mol XCl₃

\(\Rightarrow\) M_{XCl\(_3\)} = \(\dfrac{m_{XCl_3}}{n_{XCl_3}}\) = \(\dfrac{16,25}{0,1}\) = 162,5 (g/mol)

Co : 162,5 = X + 35,5.3 \(\Rightarrow\) _{X = 56 (g/mol)}

Vậy kim loại cần tìm là Fe .