\(a.NaCl+AgNO_3\rightarrow AgCl\downarrow+NaNO_3\\ a.........a..........a........a\left(mol\right)\\ KCl+AgNO_3\rightarrow KNO_3+AgCl\downarrow\\ b........b......b.......b\left(mol\right)\\ \rightarrow\left\{{}\begin{matrix}585a+745b=13,3\\143,5a+143,5b=2,87\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=0,01\\b=0,01\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}m_{NaCl\left(bđ\right)}=0,01.10.58,5=5,85\left(g\right)\\m_{KCl}=0,01.10.74,5=7,45\left(g\right)\end{matrix}\right.\\ C\%_{ddNaCl\left(bđ\right)}=\dfrac{5,85}{500}.100=1,17\%\\ C\%_{ddKCl\left(bđ\right)}=\dfrac{7,45}{500}.100=1,49\%\)
a)
$NaCl + AgNO_3 \to AgCl + NaNO_3$
$KCl + AgNO_3 \to AgCl + KNO_3$
1/10 dung dịch A phản ứng $AgNO_3$ tạo 2,87 gam kết tủa
Suy ra : dung dịch A phản ứng $AgNO_3$ tạo 28,7 gam kết tủa
Gọi $n_{NaCl} =a (mol) ; n_{KCl} = b(mol) \Rightarrow 58,5a + 74,5b = 13,3(1)$
$n_{AgCl} = a + b = \dfrac{28,7}{143,5} = 0,2(2)$
Từ (1)(2) suy ra a = b = 0,1
$m_{NaCl} = 0,1.58,5 = 5,85(gam)$
$m_{KCl} = 74,5.0,1 = 7,45(gam)$
b)
$C\%_{NaCl} = \dfrac{5,85}{500}.100\% = 1,17\%$
$C\%_{KCl} = \dfrac{7,45}{500}.100\% = 1,49\%$
a) \(NaCl+AgNO_3\rightarrow NaNO_3+AgCl\)
\(KCl+AgNO_3\rightarrow KNO_3+AgCl\)
\(n_{AgCl}=\dfrac{2,87}{143,5}=0,02\left(mol\right)\)
\(C\%_A=\dfrac{13,3}{500}.100=2,66\%\)
Lấy 1/10 dung dịch A => \(m_{\left(NaCl+KCl\right)}=2,66.50=1,33\left(g\right)\)
Gọi x, y lần lượt là số mol NaCl, KCl. Ta có hệ :
\(\left\{{}\begin{matrix}58,5x+74,5y=1,33\\x+y=0,02\end{matrix}\right.\)
=>x=0,01; y=0,01
Lấy 1/10 dung dịch A ------------> 0,01 mol NaCl, 0,01 mol KCl
=> Trong 1 dung dịch A ------------> 0,1 mol NaCl, 0,1 mol KCl
Vậy : \(m_{NaCl}=5,85\left(g\right);m_{KCl}=7,45\left(g\right)\)
b) \(C\%_{NaCl}=\dfrac{5,85}{500}.100=1,17\%\)
\(C\%_{KCl}=\dfrac{7,45}{500}.100=1,49\%\)