Question

Hòa tan 50g dd NaOH 8% vào 75g dd ZnCl2 13,6%

a) Viết PTHH

b) Tính C% dd sau PƯ

Answer 1

a)

2NaOH + ZnCl₂ --> Zn(OH)₂\(\downarrow\) + 2NaCl

b)

\(n_{NaOH}=\dfrac{50.8\%}{40}=0,1\left(mol\right)\)

\(n_{ZnCl_2}=\dfrac{75.13,6\%}{136}=0,075\left(mol\right)\)

PTHH: 2NaOH + ZnCl₂ --> Zn(OH)₂\(\downarrow\) + 2NaCl

Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{0,075}{1}\) => NaOH hết, ZnCl₂ dư

PTHH: 2NaOH + ZnCl₂ --> Zn(OH)₂\(\downarrow\) + 2NaCl

               0,1---->0,05------->0,05------>0,1

m_{dd sau pư} = 50 + 75 - 0,05.99 = 120,05 (g)

\(\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{0,1.58,5}{120,05}.100\%=4,873\%\\C\%_{ZnCl_2\left(dư\right)}=\dfrac{\left(0,075-0,05\right).136}{120,05}.100\%=2,832\%\end{matrix}\right.\)