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

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

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

\(d,\) 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)

$K_2SO_4 + BaCl_2 \to BaSO_4 + 2KCl$

b)

$n_{K_2SO_4} = 0,2.2 = 0,4(mol)$
$n_{BaCl_2} = 0,3.1 = 0,3(mol)$

Ta thấy : 

$n_{K_2SO_4} : 1 > n_{BaCl_2} : 1$ nên $K_2SO_4$ dư

$n_{BaSO_4} = n_{BaCl_2} = 0,3(mol)$
$m_{BaSO_4} = 0,3.233 = 69,9(gam)$

c) $n_{K_2SO_4} = 0,4 - 0,3 = 0,1(mol)$

$V_{dd\ sau\ pư} = 0,2 + 0,3 = 0,5(lít)$

$C_{M_{K_2SO_4} } = \dfrac{0,1}{0,5} = 0,2M$
$C_{M_{KCl}} = \dfrac{0,6}{0,5} = 1,2M$

\(a,PTHH:3NaOH+FeCl_3\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\\ 2Fe\left(OH\right)_3\rightarrow^{t^o}Fe_2O_3+3H_2O\uparrow\\ b,n_{FeCl_3}=1,5\cdot0,2=0,3\left(mol\right)\\ \Rightarrow n_{NaOH}=3n_{FeCl_3}=0,9\left(mol\right)\\ \Rightarrow V_{dd_{NaOH}}=\dfrac{0,9}{2}=0,45\left(l\right)\)

Theo đề: \(\left\{{}\begin{matrix}X:Fe\left(OH\right)_3\\A:NaCl\\Y:Fe_2O_3\end{matrix}\right.\)

Theo PT: \(n_{NaCl}=3n_{FeCl_3}=0,9\left(mol\right)\)

\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,9}{0,45+0,2}\approx1,4M\)

\(c,\) Theo PT: \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=0,3\left(mol\right);n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_X=m_{Fe\left(OH\right)_3}=0,3\cdot107=32,1\left(g\right)\\m_Y=m_{Fe_2O_3}=0,15\cdot160=24\left(g\right)\end{matrix}\right.\)

\(\begin{cases} m_{NaOH}=\dfrac{150.20\%}{100\%}=30(g)\\ m_{MgCl_2}=\dfrac{80.59,375\%}{100\%}=47,5(g) \end{cases} \Rightarrow \begin{cases} n_{NaOH}=\dfrac{30}{40}=0,75(mol)\\ n_{MgCl_2}=\dfrac{47,5}{95}=0,5(mol) \end{cases}\\ PTHH:2NaOH+MgCl_2\to Mg(OH)_2\downarrow+2NaCl\\ \text {Vì }\dfrac{n_{NaOH}}{2}<\dfrac{n_{MgCl_2}}{1} \text {nên }MgCl_2 \text { dư}\\ a,n_{Mg(OH)_2}=\dfrac{1}{2}n_{NaOH}=0,375(mol)\\ \Rightarrow m_{Mg(OH)_2}=0,375.58=21,75(g)\\ b,n_{NaCl}=m_{NaOH}=0,75(mol)\\ \Rightarrow m_{CT_{NaCl}}=0,75.58,5=43,875(g)\\ m_{dd_{NaCl}}=150+80-21,75=208,25(g)\\ \Rightarrow C\%_{NaCl}=\dfrac{43,875}{208,25}.100\%\approx 21,07\%\)

\(n_{MgCl_2}=0,15.0,2=0,03(mol)\\ PTHH:MgCl_2+2NaOH\to Mg(OH)_2\downarrow +2NaCl\\ a,n_{Mg(OH)_2}=n_{MgCl_2}=0,03(mol)\\ \Rightarrow m_{\downarrow}=m_{Mg(OH)_2}=0,03.58=1,74(g)\\ b,n_{NaOH}=2n_{MgCl_2}=0,06(mol)\\ \Rightarrow C_{M_{NaOH}}=\dfrac{0,06}{0,3}=0,2M\\ c,PTHH:Mg(OH)_2\xrightarrow{t^o}MgO+H_2O\\ \Rightarrow n_{MgO}=n_{Mg(OH)_2}=0,03(mol)\\ \Rightarrow m_{A}=m_{MgO}=0,03.40=1,2(g)\)

Ta có: \(n_{Cu\left(NO_3\right)_2}=0,2.1,5=0,3\left(mol\right)\)

PT: \(Cu\left(NO_3\right)_2+2NaOH\rightarrow Cu\left(OH\right)_{2\downarrow}+2NaNO_3\)

_______0,3_______0,6_______0,3_________0,6 (mol)

a, m_Cu(OH)2 = 0,3.98 = 29,4 (g)

b, \(V_{ddNaOH}=\dfrac{0,6}{2}=0,3\left(l\right)\)

c, \(C_{M_{NaNO_3}}=\dfrac{0,6}{0,2+0,3}=1,2M\)

Bạn tham khảo nhé!

a) \(n_{Cu\left(NO_3\right)_2}=1,5.0,2=0,3\left(mol\right)\)

\(Cu\left(NO_3\right)_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\)

\(n_{Cu\left(OH\right)_2}=n_{Cu\left(NO_3\right)_2}=0,3\left(mol\right)\)

=> \(m_{Cu\left(OH\right)_2}=29,4\left(g\right)\)

b) \(n_{NaOH}=2n_{Cu\left(OH\right)_2}=0,6\left(mol\right)\)

=> \(V_{NaOH}=\dfrac{0,6}{2}=0,3\left(l\right)\)

c) \(CM_{NaCl}=\dfrac{0,3.2}{0,2+0,3}=1,2M\)

\(n_{FeCl_3}=\dfrac{48,75}{162,5}=0,3(mol)\\ 3NaOH+FeCl_3\to Fe(OH)_3\downarrow+3NaCl\\ \Rightarrow n_{Fe(OH)_3}=0,3(mol);n_{NaOH}=n_{NaCl}=0,9(mol)\\ a,m_{Fe(OH)_3}=0,3.107=32,1(g)\\ b,m_{dd_{NaOH}}=\dfrac{0,9.40}{10\%}=360(g)\\ c,C\%_{NaCl}=\dfrac{0,9.58,5}{360+48,75-32,1}.100\%=13,98\%\\ \)

\(d,2Fe(OH)_3+3H_2SO_4\to Fe_2(SO_4)_3+6H_2O\\ \Rightarrow n_{H_2SO_4}=0,45(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,45.98}{20\%}=220,5(g)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{220,5}{1,14}=193,42(ml)\)