a. PTHH: \(CuSO_4+2NaOH--->Na_2SO_4+Cu\left(OH\right)_2\downarrow\)

b. Đổi 100ml = 0,1 lít

Ta có: \(n_{Cu\left(OH\right)_2}=\dfrac{9,8}{98}=0,1\left(mol\right)\)

Theo PT: \(n_{CuSO_4}=n_{Cu\left(OH\right)_2}=0,1\left(mol\right)\)

=> \(m_{CuSO_4}=0,1.160=16\left(g\right)\)

c. Theo PT: \(n_{NaOH}=2.n_{CuSO_4}=2.0,1=0,2\left(mol\right)\)

=> \(C_{M_{NaOH}}=\dfrac{0,2}{0,1}=2M\)

\(MgCO_3+2HCl\rightarrow MgCl_2+H_2O+CO_2\)

\(CO_2+Na\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

b. \(n_{MgCO_3}=\dfrac{21}{84}=0,25mol\) \(\Rightarrow n_{HCl}=2.0,25=0,5mol\)

\(V_{ddHCl}=\dfrac{0,5}{2}=0,25l\)

c. \(n_{CO_2}=n_{MgCO_3}=0,25mol\)

\(n_{CaCO_3}=n_{CO_2}=0,25mol\)

\(\Rightarrow m_{CaCO_3}=0,25.100=25g\)

a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_{2\downarrow}\)

\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)

b, Ta có: \(n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Mg}=0,8\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,8}{0,2}=4\left(M\right)\)

c, Theo PT: \(n_{MgO}=n_{Mg}=0,4\left(mol\right)\)

\(\Rightarrow m_{MgO}=0,4.40=16\left(g\right)\)

a)

$AgNO_3 + HCl \to AgCl + H_2O$
$NaOH + HCl \to NaCl + H_2O$
$n_{AgCl} = n_{AgNO_3} = 0,05.2 = 0,1(mol)$
$n_{AgCl} = 0,1.143,5 = 14,35(gam)$

b) $n_{HCl\ dư} = n_{NaOH} = 0,1(mol) ; n_{HCl\ pư} = n_{AgNO_3} = 0,1(mol)$

$\Rightarrow n_{HCl\ đã\ dùng} = 0,2(mol)$

$C\%_{HCl} = \dfrac{0,2.36,5}{36,5}.100\% = 20\%$

a, \(2KOH+MgSO_4\rightarrow K_2SO_4+Mg\left(OH\right)_2\)

b, Ta có: \(n_{KOH}=0,2.1=0,2\left(mol\right)\)

Theo PT: \(n_{MgSO_4}=n_{K_2SO_4}=n_{Mg\left(OH\right)_2}=\dfrac{1}{2}n_{KOH}=0,1\left(mol\right)\)

\(\Rightarrow m_{Mg\left(OH\right)_2}=0,1.58=5,8\left(g\right)\)

c, \(V_{MgSO_4}=\dfrac{0,1}{2}=0,05\left(l\right)\)

d, \(C_{M_{K_2SO_4}}=\dfrac{0,1}{0,2+0,05}=0,4\left(M\right)\)

a)

$AgNO_3 + HCl \to AgCl + HNO_3$
Theo PTHH : 

$n_{AgCl} = n_{HCl} = n_{AgNO_3} = \dfrac{340.10\%}{170} =0,2(mol)$

$m_{dd\ HCl} = \dfrac{0,2.36,5}{7,3\%} = 100(gam)$

b)

$m_{AgCl} = 0,2.143,5 = 28,7(gam)$

c)

$m_{dd\ sau\ pư} = 340 + 100 -28,7 = 411,3(gam)$

$n_{HNO_3} = n_{AgNO_3} = 0,2(mol)$
$\Rightarrow C\%_{HNO_3} = \dfrac{0,2.63}{411,3}.100\% = 3,06\%$

a) NaOH + HCl --> NaCl + H2O

b) Số mol NaOH là: n(NaOH) = 0,15mol

Theo pthh thì số mol NaCl là: n(NaCl) = 0,15mol

Khối lượng NaCl là: 0,15.58,5 = 8,775g

c)theo pthh thì số mol HCl là: 0,15mol

V(ddHCl)=100ml=0,1l

Nồng độ mol ddHCl là:

0,15/0,1=1,5M

\(n_{H_2SO_4}=1.0,2=0,2(mol)\\ H_2SO_4+BaCl_2\to BaSO_4\downarrow+2HCl\\ \Rightarrow n_{BaSO_4}=n_{BaCl_2}=0,2(mol)\\ a,m_{BaSO_4}=0,2.233=46,6(g)\\ b,V_{dd_{BaCl_2}}=\dfrac{0,2}{1,5}\approx 0,13(l)\\ c,n_{HCl}=0,4(mol)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,2+0,13}\approx 1,21M\)

\(d,\) Dd sau p/ứ là HCl nên làm quỳ tím hóa đỏ

\(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\\ H_2SO_4+BaCl_2\rightarrow BaSO_4+2HCl\\ n_{BaCl_2}=n_{BaSO_4}=n_{H_2SO_4}=0,2\left(mol\right)\\ n_{HCl}=2.0,2=0,4\left(mol\right)\\ a,m_{\downarrow}=m_{BaSO_4}=0,2.233=46,6\left(g\right)\\ b,V_{\text{dd}BaCl_2}=\dfrac{0,2}{1,5}=\dfrac{2}{15}\left(l\right)\\ c,C_{M\text{dd}HCl}=\dfrac{0,4}{\dfrac{2}{15}+0,2}=1,2\left(M\right)\\ d,V\text{ì}.c\text{ó}.\text{dd}.HCl\Rightarrow Qu\text{ỳ}.ho\text{á}.\text{đ}\text{ỏ}\)

200ml = 0,2l

\(n_{H2SO4}=1.0,2=0,2\left(mol\right)\)

Pt : \(BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl|\)

           1           1                  1              2

          0,2        0,2               0,2           0,4

a) \(n_{BaSO4}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{BaSO4}=0,2.233=46,6\left(g\right)\)

b) \(n_{BaCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(V_{ddBaCl2}=\dfrac{0,2}{1,5}=\dfrac{2}{15}\left(l\right)=133,33\left(ml\right)\)

c) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)

\(V_{ddspu}=0,2+\dfrac{2}{15}=\dfrac{1}{3}\left(l\right)\)

\(C_{M_{ddHCl}}=\dfrac{0,4}{\dfrac{1}{3}}=1,2\left(M\right)\)

d) Cho quỳ tím vào dung dịch sau phản ứng , quỳ tím hóa đỏ. Vì dung dịch thu được sau phản ứng có HCl là axit

 Chúc bạn học tốt

a) $CuSO_4 + 2NaOH \to Cu(OH)_2 + Na_2SO_4$

b) $n_{Cu(OH)_2} = n_{CuSO_4} = \dfrac{16}{160} = 0,1(mol)$
$m_{Cu(OH)_2} = 0,1.98 = 9,8(gam)$

c) $n_{NaOH} = 2n_{CuSO_4} = 0,2(mol) \Rightarrow C_{M_{NaOH}} = \dfrac{0,2}{0,1} = 2M$

PTHH: \(CuSO_4+2NaOH\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)

Ta có: \(n_{CuSO_4}=\dfrac{16}{160}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{NaOH}=0,2\left(mol\right)\\n_{Cu\left(OH\right)_2}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{NaOH}}=\dfrac{0,2}{0,1}=2\left(M\right)\\m_{Cu\left(OH\right)_2}=0,1\cdot98=9,8\left(g\right)\end{matrix}\right.\)