a)Ba+2H2O-->Ba(OH)2+H2

Ta có

n H2=3,36/22,4=0,15(mol)

Theo pthh

n H2=n Ba=0,15(mol)

m Ba=0,15.137=20,55(g)

b) Theo pthh

n Ba(OH)2=n H2=0,15(mol)

CM Ba(OH02=0,15/0,5=0,3(M)

c) Ba(OH)2+H2SO4-->BaSO4+2H2O

Ta có

n H2SO4=0,3.0,3=0,09(mol)

-->H2SO4 hết..Ba(OH)2 dư

Theo pthh

n BaSO4=n H2SO4=0,09(mol)

m BaSO4=0,09.233=20,97(g)

a, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

____0,1_____0,2______0,1_____0,1 (mol)

\(V_{H_2}=0,1.24,79=2,479\left(l\right)\)

\(m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)

b, \(C_{M_{HCl}}=\dfrac{0,2}{0,1}=1\left(M\right)\)

c, \(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)

PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Xét tỉ lệ: \(\dfrac{0,15}{1}>\dfrac{0,1}{1}\), ta được CuO dư.

Theo PT: \(n_{CuO\left(pư\right)}=n_{Cu}=n_{H_2}=0,1\left(mol\right)\Rightarrow n_{CuO\left(dư\right)}=0,05\left(mol\right)\)

⇒ m _{chất rắn} = m_{CuO (dư)} + m_Cu = 0,05.80 + 0,1.64 = 10,4 (g)

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

b, Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,1.65}{10}.100\%=65\%\\\%m_{ZnO}=35\%\end{matrix}\right.\)

c, \(n_{HCl}=0,1.0,5=0,05\left(mol\right)\)

\(n_{NaOH}=0,03.1=0,03\left(mol\right)\)

PT: \(HCl+NaOH\rightarrow NaCl+H_2O\)

Xét tỉ lệ: \(\dfrac{0,05}{1}>\dfrac{0,03}{1}\), ta được HCl dư.

→ Quỳ tím chuyển đỏ do acid dư.

Đề có thiếu gì không em nhỉ ? khối lượng riêng của HCl ? 

nHCl=0,1.5=0,5(mol)

PTHH: Zn +  2HCl -> ZnCl2 + H2

nZn=nZnCl2=nH2=nHCl/2=0,5/2=0,25(mol)

mZnCl2=0,25.136=34(g)

Mà em ơi không cho khối lượng riêng dd HCl à e,?

Câu 1 :

a)  PTHH : 

 \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\) (1) 

  \(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)(2)

b) Ta có : \(n_{Zn}=\frac{3,5}{65}\approx0,054\left(mol\right)\)

Theo phương trình hóa học (1) :

\(n_{H_2}=n_{Zn}\approx0,054\left(mol\right)\)

\(\Rightarrow V_{H_2}\approx0,054\cdot22,4=1,2096\left(l\right)\)

c) Theo phương trình hóa học (2), ta có:

\(n_{Cu}=n_{H_2}\approx0,054\left(mol\right)\)

\(\Rightarrow m_{Cu}\approx0,054\cdot64=3,456\left(g\right)\)

Bài 2:

a) Ta có : \(n_{Al}=\frac{2,7}{27}=0,1\left(mol\right)\); \(n_{HCl}=\frac{200\cdot7,3}{100\cdot36,5}=0,4\left(mol\right)\)

PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Theo phương trình hóa học : \(n_{H_2}=\frac{3}{2}n_{Al}=\frac{3}{2}.0,1=0,15\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\)

b) Theo phương trình hóa học , ta có : \(n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=0,1\cdot133,5=13,35\left(g\right)\)

Lại có: \(m_{H_2}=0,15\cdot2=0,3\left(g\right)\)

Áp dụng định luật bảo toàn khối lượng, ta có:

\(m_{Al}+m_{dd_{HCl}}=m_{dd_{AlCl_3}}+m_{H_2}\)

\(\Leftrightarrow2,7+200=m_{dd_{AlCl_3}}+0,3\)

\(\Leftrightarrow m_{dd_{AlCl_3}}=202,4\left(g\right)\)

Vậy \(C\%_{dd_{AlCl_3}}=\frac{13,35}{202,4}\cdot100\%\approx6,6\%\)

\(a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{HCl}=2\cdot0,15=0,3\left(mol\right)\)

Vì \(\dfrac{n_{Zn}}{1}>\dfrac{n_{HCl}}{2}\) nên sau p/ứ Zn dư

\(\Rightarrow n_{Zn}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\\ \Rightarrow m_{Zn}=0,15\cdot65=9,75\\ \Rightarrow m_{Zn\left(dư\right)}=13-9,75=3,25\left(g\right)\\ c,n_{H_2}=n_{Zn}=0,15\left(mol\right)\\ \Rightarrow V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\)

\(HCl+NaHCO_3\rightarrow NaCl+CO_2+H_2O\\ N_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ m_{HCl}=n_{CO_2}=0,2\left(mol\right)\\ V_{ddHCl}=\dfrac{112}{1,12}=100\left(ml\right)=0,1\left(l\right)\\ \Rightarrow C_{MddHCl}=\dfrac{0,2}{0,1}=2\left(M\right)\\ m_{HCl}=0,2.36,5=7,3\left(g\right)\\ C\%_{ddHCl}=\dfrac{7,3}{112}.100\approx6,518\%\)

\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: NaHCO₃ + HCl --> NaCl + CO₂ + H₂O

                              0,2<------------0,2

=> \(C\%\left(HCl\right)=\dfrac{0,2.36,5}{112}.100\%=6,518\%\)

\(V_{ddHCl}=\dfrac{112}{1,12}=100\left(ml\right)=0,1\left(l\right)\)

=> \(C_M\left(HCl\right)=\dfrac{0,2}{0,1}=2M\)

$n_{Ba^{2+}} = 0,1.0,5 = 0,05 < n_{SO_4^{2-}} = 0,1$ nên $SO_4^{2-}$ dư

$n_{BaSO_4} = n_{Ba^{2+}} = 0,05(mol)$
$m_{BaSO_4} = 0,05.233 = 11,65(gam)$

$n_{OH^-} = 0,1.0,5.2 + 0,1.0,5 = 0,15(mol)$
$n_{H^+} = 0,1.2 = 0,2(mol)$

$H^+ + OH^- \to H_2O$

$n_{H^+\ dư} = 0,2 - 0,15 = 0,05(mol)$

$V_{dd} = 0,1 + 0,1 + 0,1 = 0,3(lít)$

$[H^+] = \dfrac{0,05}{0,3} = \dfrac{1}{6}M$
$pH = -log( \dfrac{1}{6} ) = 0,778$

\(n_{Ba^{2+}}=0.1\cdot0.5=0.05\left(mol\right)\)

\(n_{OH^-}=0.1\cdot0.5\cdot2+0.1\cdot0.5=0.15\left(mol\right)\)

\(n_{H^+}=2\cdot0.1\cdot1=0.2\left(mol\right)\)

\(n_{SO_4^{2-}}=0.1\left(mol\right)\)

\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\)

\(0.05.........0.05.............0.05\)

\(SO_4^{2-}dư\)

\(m_{\downarrow}=0.05\cdot233=11.65\left(g\right)\)

\(H^++OH^-\rightarrow H_2O\)

\(0.15.......0.15\)

\(n_{H^+\left(dư\right)}=0.2-0.15=0.05\left(mol\right)\)

\(\left[H^+\right]=\dfrac{0.05}{0.1+0.1+0.1}=\dfrac{1}{6}\)

\(pH=-log\left(\dfrac{1}{6}\right)=0.77\)