Đáp án + Giải thích các bước giải:

\(\begin{array} {l} a)\\ Mg+2HCl\to MgCl_2+H_2\\ b)\\ n_{Mg}=\dfrac{7,2}{24}=0,3(mol)\\ n_{HCl}=\dfrac{127,75.20\%}{36,5}=0,7(mol)\\ \text{Vì }n_{Mg}<\dfrac{n_{HCl}}{2}\to HCl\text{ dư}\\ \text{Theo PT: }n_{HCl(pứ)}=2n_{Mg}=0,6(mol)\\ \to n_{HCl(dư)}=0,7-0,6=0,1(mol)\\ \to m_{HCl(dư)}=0,1.36,5=3,65(g)\\ c)\\ \text{Theo PT: }n_{H_2}=n_{Mg}=0,3(mol)\\ \to V_{H_2(đktc)}=0,3.22,4=6,72(l)\\ d)\\ \text{Theo PT: }n_{MgCl_2}=n_{Mg}=0,3(mol)\\ m_{dd\,spu}=7,2+127,75-0,3.2=134,35(g)\\ \to \begin{cases} C\%_{MgCl_2}=\dfrac{0,3.95}{134,35}.100\%≈21,21\%\\ C\%_{HCl(dư)}=\dfrac{3,65}{134,35}.100\%≈2,72\% \end{cases}\end{array}\)

a, $Mg_{}$ $+ 2HCl_{}$ → $MgCl_{2}$ $ + $H_{2}$    (1)

b, Ta có: $n_{Mg}$ = $\frac{7,2}{24}$ = 0,3 (mol)

$m_{HCl}$ = $\frac{127,75.20}{100}$ = 25,55 (mol)

⇒ $n_{HCl}$ = $\frac{25,55}{36,5}$ = 0,7 (mol)

Xét thấy: $\frac{n_{Mg}}{1}$ = 0,3 < $\frac{n_{HCl}}{2}$ = 0,35

→ Mg hết, HCl dư

Theo (1): $n_{HCl pư}$ = 2. $n_{Mg}$ = 2. 0,3 = 0,6 (mol)

→ $m_{HCl pư}$ = 0,6 . 36,6 = 21,96 (g)

→ $m_{HCl dư}$ = 25,55 - 21,96 = 3,59 (g)

Vậy HCl dư và dư 2,59 gam

c, Theo (1): $n_{H_{2}}$ = $n_{Mg}$ = 0,3 (mol)

→ $V_{H_{2}}$ = 0,3 . 22,4 = 6,72 (lít)

d, $m_{H_{2}}$ = 0,3 . 2 = 0,6 (g)

→ $m_{dd sau pư}$ = $m_{Mg}$ + $m_{dd HCl }$ - $m_{H_{2}}$ = 7,2 + 127,75 - 0,6 = 134, 35(g)

Theo (1): $n_{MgCl_{2}}$ = $n_{Mg}$ = 0,3 (mol)

→ $m_{MgCl_{2}}$ = 0,3. 95 = 28,5 (g)

C%(HCl dư) = $\frac{m_{HCl dư}}{m_{dd }}$ . 100% = $\frac{3,59}{134,35}$ . 100% = 2,67%

C%($MgCl_{2}$) = $\frac{m_{MgCl_{2}}}{m_{dd }}$ . 100% = $\frac{28,5}{134,35}$ . 100% = 21,213 %