a,
- TH1: Giả sử X là F, Y là Cl.
$\Rightarrow n_{NaCl}=n_{AgCl\downarrow}=\dfrac{57,34}{143,5}=0,4 mol$
$\Rightarrow m_{NaCl}=0,4.58,5=23,4g < 31,84g$ (TM)
- TH2: X, Y đều tạo kết tủa. Gọi chung là R.
$n_{NaR}=n_{AgR}$
$\Rightarrow \dfrac{31,84}{23+R}=\dfrac{57,34}{108+R}$
$\Rightarrow 57,34(23+R)=31,84(108+R)$
$\Leftrightarrow R=83,1$ (Br, I)
Vậy 2 muối là NaBr (x mol), NaI (y mol)
b,
- TH1:
$m_{NaF}=31,84-23,4=8,44g$
- TH2:
Ta có: $103x+150y=31,84$ (1)
$m_{\downarrow}=57,34g$
$\Rightarrow 188x+235y=57,34$ (2)
(1)(2)$\Rightarrow x=0,28; y=0,02$
$m_{NaBr}=0,28.103=28,84g$
$m_{NaI}=0,02.150=3g$