Đáp án:

 

Giải thích các bước giải:

\(\begin{array}{l} \text{Bài 1:}\\ a,\ PTHH:Mg+2HCl\to MgCl_2+H_2↑\\ n_{HCl}=\dfrac{200\times 25\%}{36,5}=1,37\ mol.\\ Theo\ pt:\ n_{Mg}=n_{MgCl_2}=n_{H_2}=\dfrac12n_{HCl}=0,685\ mol.\\ ⇒m_{Mg}=0,685\times 24=16,44\ g.\\ b,\ m_{MgCl_2}=0,685\times 95=65,075\ g.\\ c,\ m_{\text{dd spư}}=16,44+200-0,685\times 2=215,07\ g.\\ ⇒C\%_{MgCl_2}=\dfrac{65,075}{215,07}\times 100\%=30,26\%\\ \text{Bài 2:}\\ a,\ PTHH:Fe+2HCl\to FeCl_2+H_2↑\\ b,\ n_{FeCl_2}=\dfrac{63,5}{127}=0,5\ mol.\\ Theo\ pt:\ n_{Fe}=n_{H_2}=n_{FeCl_2}=0,5\ mol.\\ ⇒m=m_{Fe}=0,5\times 56=28\ g.\\ V=V_{H_2}=0,5\times 22,4=11,2\ lít.\\ Theo\ pt:\ n_{HCl}=2n_{FeCl_2}=1\ mol.\\ ⇒m_1=m_{\text{dd HCl}}={1\times 36,5}{25\%}=146\ g.\\ \text{Bài 3:}\\ a,\ PTHH:2Al+6HCl\to 2AlCl_3+3H_2↑\\ b,\ n_{AlCl_3}=\dfrac{26,7}{133,5}=0,2\ mol.\\ Theo\ pt:\ n_{Al}=n_{AlCl_3}=0,2\ mol.\\ ⇒m=m_{Al}=0,2\times 27=5,4\ g.\\ Theo\ pt:\ n_{HCl}=3n_{AlCl_3}=0,6\ mol.\\ ⇒m_1=m_{\text{dd HCl}}=\dfrac{0,6\times 36,5}{25\%}=87,6\ g.\\ Theo\ pt:\ n_{H_2}=\dfrac32n_{AlCl_3}=0,3\ mol.\\ ⇒V=V_{H_2}=0,3\times 22,4=6,72\ lít.\end{array}\)

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Bài 1:

a.

m ct HCl=$\frac{25.200}{100}$=50 g

n HCl=$\frac{50}{36,5}$ ≈1,37 mol

  Mg + 2HCl → MgCl2 + H2↑

0,685←1,37 → 0,685     0,685         mol

b.

m muối=m MgCl2=0,685.95=65,075 g

c.

-Dd sau pứ:MgCl2

mdd sau pứ=m Mg+m dd HCl-m H2

                     =0,685.24+200-0,685.2=215,07 g

C% MgCl2=$\frac{65,075}{215,07}$.100% ≈30,26 %

Bài 2:

a.

n FeCl2=$\frac{63,5}{127}$=0,5 mol

Fe + 2HCl → FeCl2 + H2↑

0,5    1        ←0,5  →   0,5         mol

b.

m=m Fe=0,5.56=28 g

V =V H2(đktc)=0,5.22,4=11,2 l

m ct HCl=1.36,5=36,5 g

m1=m dd HCl=$\frac{36,5.100}{25}$=146 g

Bài 3:

a.

n AlCl3=$\frac{26,7}{133,5}$=0,2 mol

2Al+6HCl→2AlCl3+3H2↑

0,2    0,6  ← 0,2  →   0,3      mol

b.

m=m Al=0,2.27=5,4 g

V=V H2(đktc)=0,3.22,4=6,72 l

m ct HCl=0,6.36,5=21,9 g

m1=m dd HCl=$\frac{21,9.100}{25}$=87,6 g

---------------------Nguyễn Hoạt------------------------