a)ko pứ

b) ko pứ

c)2Al+3Zn(NO3)2->2Al(NO3)3+3Zn

# quy ước theo mức độ hoạt động của dãy kim loại

nè xl nhé

Yêu cầu đề bài

cách viết sau chỉ ý gì?

a) \(\left\{{}\begin{matrix}m_{HCl}=500.10\%+500.20\%=150\left(g\right)\\m_{dd}=500+500=1000\left(g\right)\end{matrix}\right.\)

`=>` \(C\%_{HCl\left(mới\right)}=\dfrac{150}{1000}.100\%=15\%\)

b) \(\left\{{}\begin{matrix}n_{HCl}=\dfrac{150}{36,5}=\dfrac{300}{73}\left(mol\right)\\V_{dd}=\dfrac{1000}{1,2}=\dfrac{2500}{3}\left(ml\right)=\dfrac{5}{6}\left(l\right)\end{matrix}\right.\)

`=>` \(C_{M\left(HCl\right)}=\dfrac{\dfrac{300}{73}}{\dfrac{5}{6}}=\dfrac{360}{73}M\)

\(\begin{array}{l}
a)\\
{m_{HCl}} = 500 \times 10\%  + 500 \times 20\%  = 150g\\
{m_{{\rm{dd}}}} = 500 + 500 = 1000g\\
{C_\% }HCl = \dfrac{{150}}{{1000}} \times 100\%  = 15\% \\
b)\\
{V_{{\rm{dd}}HCl}} = \dfrac{{1000}}{{1,2}} = 833,33\,ml = 0,83333l\\
{n_{HCl}} = \dfrac{{150}}{{36,5}} = 4,1\,mol\\
{C_M}HCl = \dfrac{{4,1}}{{0,83333}} = 4,92M
\end{array}\)

a) PTHH: `Cu + 2AgNO_3 -> Cu(NO_3)_2 + 2Ag`

b) Gọi `n_{Cu(pư)} = a (mol)`

Theo PT: `n_{Ag} = 2n_{Cu} = 2a (mol)`

Ta có: `m_{tăng} = m_{Ag} - m_{Cu} = 2a.108 - 64a = 13,2 - 6`

`=>` \(a=\dfrac{9}{190}\left(mol\right)\)

`=>` \(m_{Cu\left(pư\right)}=\dfrac{9.64}{190}=\dfrac{288}{95}\left(g\right)\)

Ta có: \(\left\{{}\begin{matrix}n_{H_2SO_4}=0,4.0,0375=0,015\left(mol\right)\\n_{HCl}=0,4.0,0125=0,005\left(mol\right)\end{matrix}\right.\)

`=>` \(n_{H^+}=0,015.2+0,005=0,035\left(mol\right)\)

`=>` \(\left[H^+\right]=\dfrac{0,035}{0,4}=0,0875M\)

\(n_{H_2SO_4}=V\cdot CM_{H_2SO_4}=0.4\cdot0.0375=0.015mol\\ n_{HCl}=V\cdot CM_{HCl}=0.4\cdot0.0125=0.005mol\\ \left[H^+\right]=\dfrac{n_H}{V}=\dfrac{\left(0.015\cdot2\right)+0.005}{0.4}=0.0875\left(\dfrac{mol}{l}\right)\)

a) \(n_{KOH}=\dfrac{160.21\%}{56}=0,6\left(mol\right)\)
PTHH: \(Fe_2\left(SO_4\right)_3+6KOH\rightarrow2Fe\left(OH\right)_3\downarrow+3K_2SO_4\)

                 0,1<--------0,6--------->0,2

`=>` \(\left\{{}\begin{matrix}x=0,1.400=40\left(g\right)\\a=0,2.107=21,4\left(g\right)\end{matrix}\right.\)

\(n_{SO_4^{2-}}=0,5.0,9=0,45\left(mol\right)\)

Ta có: \(n_{SO_4^{2-}}=3n_{Fe_2\left(SO_4\right)_3}+n_{Na_2SO_4}\Rightarrow n_{Na_2SO_4}=0,45-0,1.3=0,15\left(mol\right)\)

`=>` \(y=0,15.142=21,3\left(g\right)\)

b) PT ion rút gọn: \(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\downarrow\)

                                           0,45----->0,45

`=>` \(m_{kt}=0,45.233=104,85\left(g\right)\)

Đặt \(n_{Zn}=x\left(mol\right)\)

\(Zn+CuSO_4\rightarrow ZnSO_4+Cu\downarrow\)

 x           x                                 x       ( mol )

\(m_{giảm}=m_{Zn}-m_{Cu}\)

\(\Leftrightarrow65x-64x=25-24,96\)

\(\Leftrightarrow x=0,04\left(mol\right)\)

\(m_{Zn}=0,04.65=2,6\left(g\right)\)

\(m_{CuSO_4}=0,04.160=6,4\left(g\right)\)

\(n_{Na}=\dfrac{5,52}{23}=0,24\left(mol\right)\)

PTHH: `2Na + 2H_2O -> 2NaOH + H_2`

Theo PT: `n_{NaOH} = n_{Na} = 0,24 (mol)`

Ta có: `n_{Ba(OH)_2} = 0,04.0,5 = 0,02 (mol)`

`=>` \(\sum n_{OH^-}=0,24+0,02.2=0,28\left(mol\right)\)

`=>` \(\left[OH^-\right]=\dfrac{0,28}{0,5}=0,56M\)

Những cặp chất xảy ra phản ứng:

a) \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

c) \(Fe+CuSO_4\rightarrow FeSO_4+Cu\)

d) \(Zn+Pb\left(NO_3\right)_2\rightarrow Zn\left(NO_3\right)_2+Pb\)