PTHH : FeCl3 + 3NaOH → Fe(OH)3 + 3NaCl
nFeCl3=\(\dfrac{26}{162,5}\)=0,16 mol . Theo tỉ lệ pt => nNaOH = 0,16.3 =0,48 mol.
<=> mNaOH = 0,48.40= 19,2 gam
C% =\(\dfrac{m_{\left(ct\right)}}{m_{\left(dd\right)}}.100\) => m dung dịch NaOH 10% = \(\dfrac{19,2.100}{10}\)= 192 gam
PT: \(FeCl_3+3NaOH\rightarrow3NaCl+Fe\left(OH\right)_{3\downarrow}\)
Ta có: \(n_{FeCl_3}=\dfrac{26}{162,5}=0,16\left(mol\right)\)
Theo PT: \(n_{NaOH}=3n_{FeCl_3}=0,48\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,48.40=19,2\left(g\right)\)
\(\Rightarrow m_{ddNaOH}=\dfrac{19,2.100}{10}=192\left(g\right)\)
Bạn tham khảo nhé!
