a) Gọi số mol Zn, Al là a,b (mol)

=> 65a + 27b = 9,2 (1)

\(n_{H_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

             a---->2a-------------->a

            2Al + 6HCl --> 2AlCl₃ + 3H₂

            b----->3b--------------->1,5b

=> a + 1,5b = 0,25 (2)

(1)(2) => a = 0,1 (mol); b = 0,1 (mol)

=> \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,1.65}{9,2}.100\%=70,65\%\\\%m_{Al}=\dfrac{0,1.27}{9,2}.100\%=29,35\%\end{matrix}\right.\)

b) n_HCl = 2a + 3b = 0,5 (mol)

=> \(C\%_{dd.HCl}=\dfrac{0,5.36,5}{200}.100\%=9,125\%\)

a)

\(SO_3 + H_2O \to H_2SO_4\)

Theo PTHH : \(n_{H_2SO_4} = n_{SO_2} = \dfrac{8}{80} = 0,1(mol)\)

\(\Rightarrow C\%_{H_2SO_4} = \dfrac{0,1.98}{200}.100\% = 4,9\%\)

b)

\(2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O\)

Theo PTHH : \(n_{NaOH} = 2n_{H_2SO_4} = 0,2(mol)\\ \Rightarrow m_{dd\ NaOH} = \dfrac{0,2.40}{4\%} = 200(gam)\)

C%NaOH=\(\dfrac{20}{200}100\)=10%

2 

mNaCl= 34g

=>C%NaCl=\(\dfrac{34}{200}.100\)=17%

3

m KOH=20g

=>C%=\(\dfrac{35}{15+200}\)=16,279%

4

C%KCl=\(\dfrac{25}{275}100\)=9,09%

a) \(C\%=\dfrac{m_{KCl}}{m_{ddKCl}}.100\%=\dfrac{10}{300}.100\%\approx3,3\%\)

b) Đổi: \(1500ml=1,5l\)

\(C_{MCuSO_4}=\dfrac{n}{V}=\dfrac{3}{1,5}=2M\)

a, \(C\%_{KCl}=\dfrac{20}{20+60}.100\%=25\%\)

b, \(C\%=\dfrac{40}{40+150}.100\%\approx21,05\%\)

c, \(C\%_{NaOH}=\dfrac{60}{60+240}.100\%=20\%\)

d, \(C\%_{NaNO_3}=\dfrac{30}{30+90}.100\%=25\%\)

e, \(m_{NaCl}=150.60\%=90\left(g\right)\)

f, \(m_{ddA}=\dfrac{25}{10\%}=250\left(g\right)\)

g, \(n_{NaOH}=120.20\%=24\left(g\right)\)

Gọi: n_{NaOH (thêm vào)} = a (g)

\(\Rightarrow\dfrac{a+24}{a+120}.100\%=25\%\Rightarrow a=8\left(g\right)\)

\(a.m_{dd}=21,6+400=421,6\left(g\right)\\ b.V_{dd}=400\left(mL\right)\\ c.C_M=\dfrac{\dfrac{21,6}{40}}{0,4}=1,35\left(mol\cdot L^{-1}\right)\)

\(a.m_A=m_{NaOH\left(loãng\right)}=21,6+400=421,6\left(g\right)\)

b,c. thiếu D nhé

\(a.NaCl+AgNO_3\rightarrow AgCl\downarrow+NaNO_3\\ a.........a..........a........a\left(mol\right)\\ KCl+AgNO_3\rightarrow KNO_3+AgCl\downarrow\\ b........b......b.......b\left(mol\right)\\ \rightarrow\left\{{}\begin{matrix}585a+745b=13,3\\143,5a+143,5b=2,87\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=0,01\\b=0,01\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}m_{NaCl\left(bđ\right)}=0,01.10.58,5=5,85\left(g\right)\\m_{KCl}=0,01.10.74,5=7,45\left(g\right)\end{matrix}\right.\\ C\%_{ddNaCl\left(bđ\right)}=\dfrac{5,85}{500}.100=1,17\%\\ C\%_{ddKCl\left(bđ\right)}=\dfrac{7,45}{500}.100=1,49\%\)

a)

$NaCl + AgNO_3 \to AgCl + NaNO_3$
$KCl + AgNO_3 \to AgCl + KNO_3$

1/10 dung dịch A phản ứng $AgNO_3$ tạo 2,87 gam kết tủa

Suy ra : dung dịch A phản ứng $AgNO_3$ tạo 28,7 gam kết tủa

Gọi $n_{NaCl} =a (mol) ; n_{KCl} = b(mol) \Rightarrow 58,5a + 74,5b = 13,3(1)$

$n_{AgCl} = a + b = \dfrac{28,7}{143,5} = 0,2(2)$
Từ (1)(2) suy ra a = b = 0,1

$m_{NaCl} = 0,1.58,5 = 5,85(gam)$
$m_{KCl} = 74,5.0,1 = 7,45(gam)$

b)

$C\%_{NaCl} = \dfrac{5,85}{500}.100\% = 1,17\%$
$C\%_{KCl} = \dfrac{7,45}{500}.100\% = 1,49\%$

a) \(NaCl+AgNO_3\rightarrow NaNO_3+AgCl\)

\(KCl+AgNO_3\rightarrow KNO_3+AgCl\)

\(n_{AgCl}=\dfrac{2,87}{143,5}=0,02\left(mol\right)\)

\(C\%_A=\dfrac{13,3}{500}.100=2,66\%\)

Lấy 1/10 dung dịch A => \(m_{\left(NaCl+KCl\right)}=2,66.50=1,33\left(g\right)\)

Gọi x, y lần lượt là số mol NaCl, KCl. Ta có hệ :

 \(\left\{{}\begin{matrix}58,5x+74,5y=1,33\\x+y=0,02\end{matrix}\right.\)

=>x=0,01; y=0,01

Lấy 1/10 dung dịch A ------------> 0,01 mol NaCl, 0,01 mol KCl

=> Trong 1 dung dịch A ------------> 0,1 mol NaCl, 0,1 mol KCl

Vậy : \(m_{NaCl}=5,85\left(g\right);m_{KCl}=7,45\left(g\right)\)

b) \(C\%_{NaCl}=\dfrac{5,85}{500}.100=1,17\%\)

\(C\%_{KCl}=\dfrac{7,45}{500}.100=1,49\%\)

a)

Khối lượng của dung dịch:

\(m_{dd}=m_{ct}+m_{dm}=20+180=200\left(g\right)\)

Nồng độ phần trăm của dung dịch:

\(C\%=\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{20}{200}.100\%=10\%\)

b) đề sai nha bạn

=50/400.100%=12,5%.

Em đăng vào môn Hoá hấy