Câu 36 :

$n_{HCl} = 0,5.5 = 2,5(mol)$

$m_{dd\ HCl} = \dfrac{2,5.36,5}{36,5\%} = 250(gam)$
$V_{dd\ HCl} = \dfrac{m}{D} = \dfrac{250}{1,19} = 210(ml)$

Đáp án A

Câu 37  :

$Zn + 2HCl \to ZnCl_2 + H_2$
Theo PTHH :

$n_{H_2} = n_{Zn} = \dfrac{13}{65} = 0,2(mol)$
$V_{H_2} = 0,2.22,4 = 4,48(lít)

Đáp án C

Câu 50 :

Gọi $n_{FeSO_4.7H_2O} = a(mol)

Sau khi pha :

$m_{dd} = 278a + 372,2(gam)$
$n_{FeSO_4} = n_{FeSO_4.7H_2O} = a(mol)$

Vậy :

$C\%_{FeSO_4} = \dfrac{152a}{278a + 372,2}.100\% = 3,8\%$

$\Rightarrow a = 0,1(mol)$
$m_{FeSO_4.7H_2O} = 0,1.278 = 27,8(gam)$

Đáp án D

nHCl=0,5.5=2,5(mol)

mdd HCl=\(\dfrac{2,5.36,5}{36,5\%}\)%=250(gam)
Vdd HCl=\(\dfrac{250}{1,19}\)=210(ml)

a) GS có x mol HCl

\(\Rightarrow m_{HCl}\)mHClHCl=36,5x

\(\Rightarrow m_{dd_{HCl}}\)=36,5x/37%=98,65x

\(\Rightarrow V_{dd}=\frac{m_{dd}}{D}\)=\(\frac{98,65x}{1,19}\)=82,9x (ml)

\(\Rightarrow CM_{dd_{HCl}}\)=x/0,0829x=12M

b) GS có x mol HCl

\(\Rightarrow m_{HCl}\)=36,5x

\(V_{dd_{HCl}}\)=x/10,81 lít

\(\Rightarrow m_{dd_{HCl}}\)=1/10,81.1000.1,17.x=108,233x

\(\Rightarrow\)C%=36,5/108,233.100%=33,72%

GS có x mol HClHCl

=>mHClHCl=36,5x

=>mdd HClHCl=36,5x/37%=98,65x

=>Vdd=mdd/D=98,65x/1,19=82,9x (ml)

=>CM dd HClHCl=x/0,0829x=12M

b) GS có x mol HClHCl

=>mHClHCl=36,5x

Vdd HClHCl=x/10,81 lít

=>mdd HClHCl=1/10,81.1000.1,17.x=108,233x

=>C%=36,5/108,233.100%=33,72%

n_HCl = 5.0,5 = 2,5 (mol)

⇒ m_HCl = 2,5.36,5 = 91,25 (g)

\(\Rightarrow m_{ddHCl}=\dfrac{91,25}{36\%}=\dfrac{9125}{36}\left(g\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{\dfrac{9125}{36}}{1,19}\approx213\left(ml\right)=0,213\left(l\right)\)

1.\(C_M=\dfrac{10.D.C\%}{M}=\dfrac{10.1,19.37}{36,5}\approx12,063M\)

2.\(C\%=\dfrac{C_M.M}{10.D}=\dfrac{10,81.36,5}{10.1,17}\approx33,724\%\)

Bài 1: Coi dd HCl đó chứa \(m=1,19g;V=1ml=0,001l\) (Phù hợp với D)

Theo bài ra ta có:

\(m_{HCl}=1,19.37\%=0,4403\left(g\right)\\ \Rightarrow n_{HCl}=\dfrac{0,4403}{36,5}=\dfrac{4403}{365000}\left(mol\right)\)

Do đó \(C_{M_{\text{dd}HCl}}=\dfrac{\dfrac{4403}{365000}}{0,001}\approx12,063M\)

Bài 2:

Coi dd HCl đó có

\(m=1,17g;V=1ml=0,001l\)

Theo gt ta có:

\(n_{HCl}=10,81.0,001=0,01081\left(mol\right)\\ \Rightarrow m_{HCl}=0,01081.36,5=0,394565\left(g\right)\)

Do đó \(\%C_{HCl}=\dfrac{0,394565.100\%}{1,17}\approx33,724\%4\%\)

\(m_{ddHCl.20\%}=2000\times1,1=2200\left(g\right)\)

\(\Rightarrow m_{HCl.20\%}=2200\times20\%=440\left(g\right)\)

Gọi x,y lần lượt là khối lượng dd của dd HCl.36% và dd HCl.12%

\(\Rightarrow m_{HCl.36\%}=x\times36\%=\frac{9}{25}x\left(g\right)\)

\(m_{HCl.12\%}=y\times12\%=\frac{3}{25}y\left(g\right)\)

Ta có: \(\left\{{}\begin{matrix}x+y=2200\\\frac{9}{25}x+\frac{3}{25}y=440\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=733,33\\y=1466,67\end{matrix}\right.\)

Vậy \(m_{ddHCl.36\%}=733,33\left(g\right)\)

\(m_{ddHCl.12\%}=1466,67\left(g\right)\)

\(V_{H_2O}=V\left(l\right)\)

\(C_{M_{HCl\left(ls\right)}}=\dfrac{1.25\cdot0.016}{0.016+V}=0.25\left(M\right)\)

\(\Rightarrow V=0.064\left(l\right)=64\left(ml\right)\)