Đáp án:

$n_{AgCl}=\frac{3,5875}{143,5}=0,025(mol)$

$RCl_x+xAgNO_3→xAgCl↓+R(NO3)_x$  

$\text{ Theo phương trình hoá học với 50 ml}$

$n_{RCl_x}=\frac{0,025}{x}(mol)$

Trong 300ml : $\frac{0,025.6}{x}=\frac{0,15}{x}$

$⇒M_R=\frac{8,025.x}{0,15}=53,5x$

$⇔R+35,5x=53,5x$

$⇒R=18x$

Với:

$x=1; R=18 (NH_4)$

$⇒R:NH_4Cl$

Em tham khảo nha:

\(\begin{array}{l}
MC{l_n} + nAgN{O_3} \to M{(N{O_3})_3} + nAgCl\\
{n_{AgCl}} = \dfrac{{3,5875}}{{143,5}} = 0,025\,mol\\
{n_{MC{l_n}}} = 0,025 \times \dfrac{1}{n} = \dfrac{1}{{40n\,}}mol\\
{n_{MC{l_n}\,trong\,X}} = \dfrac{1}{{40n}} \times 6 = \dfrac{3}{{20n}}\,mol\\
{M_{MC{l_n}}} = \dfrac{{8,025}}{{\frac{3}{{20n}}}} = 53,5n\,g/mol\\
 \Rightarrow M + 35,5n = 53,5n \Rightarrow M = 18n\,g/mol
\end{array}\)

\(\begin{array}{l}
n = 1 \Rightarrow {M_M} = 18g/mol \Rightarrow M:N{H_4}\\
CTHH:N{H_4}Cl
\end{array}\)