Đáp án:
b) 1M
c) 5,6g
d) Bạc
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,1\,mol\\
{C_M}{H_2}S{O_4} = \dfrac{{0,1}}{{0,1}} = 1M\\
c)\\
{n_{Fe}} = {n_{{H_2}}} = 0,1\,mol\\
{m_{Cu}} = 11,2 - 0,1 \times 56 = 5,6g\\
d)\\
Cu + 2RN{O_3} \to Cu{(N{O_3})_2} + 2R\\
{n_{Cu}} = \dfrac{{5,6}}{{64}} = 0,0875mol\\
{n_R} = 0,0875 \times 2 = 0,175\,mol\\
{M_R} = \dfrac{{18,9}}{{0,175}} = 108g/mol \Rightarrow R:Ag
\end{array}\)