Em tham khảo nha:
\(\begin{array}{l}
VD\,5:\\
Cu + 2AgN{O_3} \to Cu{(N{O_3})_2} + 2Ag\\
\text{ Gọi x là số mol $AgNO_3$ }\\
{n_{Cu}} = \dfrac{{{n_{AgN{O_3}}}}}{2} = 0,5x\,mol\\
{m_{Ag}} - {m_{Cu}} = 11,4 \Leftrightarrow 108x - 0,5x \times 64 = 11,4\\
\Rightarrow a = 0,15\,mol\\
{C_M}AgN{O_3} = \dfrac{{0,15}}{{0,3}} = 0,5M \Rightarrow a = 0,5\\
VD6:\\
2NaHC{O_3} + {H_2}S{O_4} \to N{a_2}S{O_4} + 2C{O_2} + 2{H_2}O\\
Ca{(HC{O_3})_2} + {H_2}S{O_4} \to CaS{O_4} + 2C{O_2} + 2{H_2}O\\
Ba{(HC{O_3})_2} + {H_2}S{O_4} \to BaS{O_4} + 2C{O_2} + 2{H_2}O\\
{n_{{H_2}O}} = {n_{C{O_2}}} = 0,4\,mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{0,4}}{2} = 0,2\,mol\\
BTKL:\\
a = 35 + 0,2 \times 98 - 0,4 \times 44 - 0,4 \times 18 = 29,8g\\
VD7:\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
\text{ Gọi x là số mol HCl }\\
{n_{{H_2}O}} = \dfrac{{{n_{HCl}}}}{2} = 0,5a\,mol\\
BTKL:28 + x \times 36,5 = 55,5 + 18 \times 0,5x \Rightarrow x = 1\\
{C_M}HCl = \dfrac{1}{{0,5}} = 2M \Rightarrow a = 2
\end{array}\)
