Gọi số mol NaCl là x mol ; NaBr là y mol
Giải hệ phương trình :
\(\left\{{}\begin{matrix}58,5x+103y=11\\143,5x+188y=23,75\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)
\(m_{NaCl}=0,1.58,5=5,85\left(g\right)\)
\(m_{NaBr}=0,05.103=5,515\left(g\right)\)
Gọi a là mol NaCl; b là mol NaBr
\(\Rightarrow58,5a+103b=11\left(1\right)\)
Bảo toàn nguyên tố:
\(n_{AgCl}=n_{NaCl}=a\left(mol\right)\)
\(n_{AgBr}=n_{NaBr}=b\left(mol\right)\)
\(\Rightarrow143,5x+188b=23,75\left(2\right)\)
\(\left(1\right)+\left(2\right)\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\)
\(m_{NaCl}=0,1.58,5=5,85\left(g\right)\)
\(m_{NaBr}=0,05.103=5,15\left(g\right)\)