Đáp án:
$V=0,56; m=2,785$
Giải thích các bước giải:
\(\begin{array}{l}
{P_1}:\\
R + {O_2} \to {R_2}{O_n}\\
BT\,e:{n_R} \times n = 4{n_{{O_2}}}(1)\\
{m_{{O_2}}} = 1,41 - \dfrac{{2,02}}{2} = 0,4\,mol\\
{n_{{O_2}}} = \dfrac{{0,4}}{{32}} = 0,0125\,mol\\
{P_2}:\\
2R + 2nHCl \to 2RC{l_n} + n{H_2}\\
BT\,e:{n_R} \times n = 2{n_{{H_2}}}(2)\\
(1),(2) \Rightarrow {n_{{H_2}}} = 2{n_{{O_2}}} = 0,025\,mol\\
{V_{{H_2}}} = 0,025 \times 22,4 = 0,56l\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,05\,mol\\
BTKL:\\
m = 1,01 + 0,05 \times 36,5 - 0,025 \times 2 = 2,785g
\end{array}\)