a. 

Chất tham gia : alunium , hydrochloric acid

Chất sản phẩm : aluminium chloride , khí hydrogen 

b. 

alunium + hydrochloric acid →  aluminium chloride + khí hydrogen 

c. 

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

d.

\(m_{H_2}=m_{Al}+m_{HCl}-m_{AlCl_2}=10.5+20-15.5=15\left(g\right)\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{HCl}=2.n_{Zn}=2.0,1=0,2\left(mol\right)\\a, \Rightarrow m_{HCl}=0,2.36,5=7,3\left(g\right)\\ b,n_{H_2}=_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow m_{ZnCl_2}=136.0,1=13,6\left(g\right)\\ c,Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\\ n_{Fe_2O_3}=\dfrac{n_{H_2}}{3}=\dfrac{0,1}{3}=\dfrac{1}{30}\left(mol\right)\\ \Rightarrow m_{Fe_2O_3}=\dfrac{1}{30}.160=\dfrac{16}{3}\left(g\right)\)

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

a: \(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)

\(\Leftrightarrow n_{ZnCl_2}=0.1\left(mol\right)\)

\(\Leftrightarrow n_{HCl}=0.2\left(mol\right)\)

\(m_{HCl}=0.2\cdot36.5=7.3\left(g\right)\)

b: \(m_{ZnCl_2}=0.1\cdot136=13.6\left(g\right)\)

A.

a: \(n_{Zn}=\dfrac{2.6}{65}=0.04\left(mol\right)\)

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

0,04                 0,04

\(m_{ZnCl_2}=0.04\left(65+35.5\cdot2\right)=5.44\left(g\right)\)

b: \(n_{HCl}=2\cdot0.04=0.08\left(mol\right)\)

\(m_{ct\left(HCl\right)}=0.08\cdot36.5=2.92\left(g\right)\)

\(C\%\left(HCl\right)=\dfrac{2.92}{200}=0.0146\)

d: \(n_{H_2}=n_{Zn}=0.04\left(mol\right)\)

V=0,04*22,4=0,896(lít)

\(a)n_{Zn}=\dfrac{2,6}{65}=0,04mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

0,04         0,08          0,04         0,04

\(m_{ZnCl_2}=0,04.136=5,44g\\ b)C_{\%HCl}=\dfrac{0,08.36,5}{200}\cdot100=1,46\%\\ c)C_{M\left(HCl\right)}=\dfrac{0,08}{0,2}=0,4M\\ d)V_{H_2}=0,04.22,4=0,896l\\ e)H_2+CuO\underrightarrow{t^0}Cu+H_2O\\ n_{Cu}=n_{H_2}=0,04mol\\ m_{Cu}=0,04.64=2,56g\)

$PTHH:Zn+2HCl\to ZnCl_2+H_2\uparrow$

$n_{Zn}=\dfrac{13}{65}=0,2(mol)$

Theo PT: $n_{ZnCl_2}=n_{H_2}=0,2(mol);n_{HCl}=0,4(mol)$

$a)m_{axit}=m_{HCl}=n.M=0,4.36,5=14,6(g)$

$b)m_{ZnCl_2}=n.M=0,2.136=27,2(g)$

$c)V_{H_2(đktc)}=n.22,4=0,2.22,4=4,48(lít)$

Số mol kẽm là : 

\(n=\dfrac{m}{M}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH : Zn + 2HCL -> ZnCl₂ + H₂

1 2 1 1 

0,2 mol -> 0,4 mol 0,2 mol 0,2 mol 

a, Khối lượng HCL là : 

\(m=n.M=0,4.35,5=14,2\left(g\right)\)

b, Khối lượng ZnCL₂ là : 

\(m=n.M=0,1.136=13,6\left(g\right)\)

c, Thể tích H₂ là : V = n . 22,4 = \(0,1.22,4=2,24\left(l\right)\)

\(m_{CaCO_3}=400\cdot90\%=360\left(g\right)\)

\(m_{trơ}=400-360=40\left(g\right)\)

\(n_{CaCO_3}=\dfrac{360}{100}=3.6\left(mol\right)\)

\(a.\)

\(n_{CaCO_3\left(pư\right)}=3.6\cdot75\%=2.7\left(mol\right)\)

\(CaCO_3\underrightarrow{^{^{t^0}}}CaO+CO_2\)

\(2.7........2.7...........2.7\)

\(m_X=m_{CaO}+m_{CaCO_3\left(dư\right)}+m_{trơ}=2.7\cdot56+\left(3.6-2.7\right)\cdot100+40=281.2\left(g\right)\)

\(b.\)

\(\%CaO=\dfrac{2.7\cdot56}{281.2}\cdot100\%=53.77\%\)

\(V_{CO_2}=2.7\cdot22.4=60.48\left(l\right)\)

\(n_{Fe}=\dfrac{11,2}{56}=0,2mol\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,2     0,4          0,2         0,2

\(V_{H_2}=0,2\cdot22,4=4,48l\)

\(m_{FeCl_2}=0,2\cdot\left(56+2\cdot35,5\right)=25,4g\)