Question

Answer 1

\(n_{KCl}=a\left(mol\right),n_{KBr}=b\left(mol\right)\)

Giả sử : 

\(a+b=1\left(mol\right)\left(1\right)\)

\(KCl+AgNO_3\rightarrow KNO_3+AgCl\)

\(KBr+AgNO_3\rightarrow KNO_3+AgBr\)

\(m_{\downarrow}=143.5a+188b=170\left(a+b\right)\left(g\right)\)

\(\Leftrightarrow26.5a-18b=0\left(2\right)\)

\(\left(1\right),\left(2\right):a=\dfrac{36}{89},b=\dfrac{53}{89}\)

\(\%KCl=\dfrac{\dfrac{36}{89}\cdot74.5\cdot100\%}{\dfrac{36}{89}\cdot74.5+\dfrac{53}{89}\cdot119}=29.84\%\)