Câu 3: (VDC) (2,0 điểm) 1...

- Hướng dẫn giải

Phương pháp giải:

1. Bảo toàn nguyên tố, bảo toàn e

2. Đặt phần 2 gấp k lần mỗi số mol của phần 1

Giải chi tiết:

1.

\(\underbrace {MgC{l_2}}_{0,13}\xrightarrow{{}}\left\{ \begin{gathered}
Mg:0,13 \hfill \\
Ca:x \hfill \\
O:y \hfill \\
\end{gathered} \right.\xrightarrow{{}}\left\{ \begin{gathered}
24.0,13 + 40x + 16y = 10,72 \hfill \\
\xrightarrow{{BT\,e}}\left\{ \begin{gathered}
2.nMg + 2nCa = 2.nO + 2n{H_2} \hfill \\
\to 2.0,13 + 2x = 2y + 0,29 \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered} \right. = > \left\{ \begin{gathered}
x = 0,14 \hfill \\
y = 0,125 \hfill \\
\end{gathered} \right. = > a = 15,54(g)\)

Vậy giá trị của a = 15,54 (g)

2. 

a.

\(m\left\{ \begin{gathered}
A{l_4}{C_3} \hfill \\
Ca{C_2} \hfill \\
\end{gathered} \right.\xrightarrow{{ + H2O}}\left\{ \begin{gathered}
C{H_4} \hfill \\
{C_2}{H_2} \hfill \\
\end{gathered} \right. \to \left[ \begin{gathered}
\xrightarrow{{ + B{r_2}:\,0,1}} \hfill \\
\xrightarrow{{ + \,{O_2}}}\left\{ \begin{gathered}
C{O_2} \hfill \\
{H_2}O \hfill \\
\end{gathered} \right.:m\,dd\,giam = 69,525\,g \hfill \\
\end{gathered} \right.\)

Al₄C₃ +12H₂O → 4Al(OH)₃ + 2CH₄↑

CaC₂ + 2H₂O → Ca(OH)₂ + CH≡CH↑

\(\begin{gathered}
{P_1}\left\{ \begin{gathered}
C{H_4}:x \hfill \\
\xrightarrow[{0,1}]{{B{r_2}}}{C_2}{H_2}:0,05 \hfill \\
\end{gathered} \right. \to {P_2}\left\{ \begin{gathered}
C{H_4}:kx \hfill \\
{C_2}{H_2}:0,05k \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
C{O_2}:xk + 0,1k \hfill \\
{H_2}O:2xk + 0,05k \hfill \\
\end{gathered} \right. \hfill \\
\xrightarrow[{69,525\,g}]{{m\,dd\,giam}}\left\{ \begin{gathered}
mBaC{O_3} - m(C{O_2} + {H_2}O) = 69,525 \hfill \\
\to 153(X + 0,1)k - 18(2x + 0,05)k = 69,525 \hfill \\
\end{gathered} \right. \to k = \frac{{69,525}}{{117x + 14,4}} \hfill \\
\end{gathered} \) 

Mặt khác:

\(\begin{gathered}
\xrightarrow[{0,5}]{{aX}}n({P_1} + {P_2}) = (x + 0,05).(k + 1) = 0,05\xrightarrow{{k = \frac{{69,525}}{{117x + 14,4}}}}\left\{ \begin{gathered}
x = 0,075 \hfill \\
k = 3 \hfill \\
\end{gathered} \right. \hfill \\
\to \left\{ \begin{gathered}
C{H_4}:0,3 \hfill \\
{C_2}{H_2}:0,2 \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
\% V(x):60\% ;40\% \hfill \\
\left\{ \begin{gathered}
A{l_4}{C_3}:0,1 \hfill \\
Ca{C_2}:0,2 \hfill \\
\end{gathered} \right. \to m = 27,2\,g \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered} \)

Vậy giá trị của m = 27,2 và %V các khí trong X là CH₄ 60%; C₂H₂ 40%

b.

\({P_2}\left\{ \begin{gathered}
C{H_4}:0,225 \hfill \\
{C_2}{H_2}:0,15 \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
{C_2}{H_2}\, + A{g_2}O\xrightarrow{{N{H_3}}}{C_2}A{g_2} \downarrow + {H_2}O \hfill \\
0,15 \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,15 \hfill \\
\end{gathered} \right. \to a = 36g\)

Vậy giá trị của a = 36 (gam)