a.b.c.\(n_{Zn}=\dfrac{m}{M}=\dfrac{13}{65}=0,2mol\)

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

0,2                     0,2        0,2  ( mol )

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

\(V_{H_2}=n.22,4=0,2.22,4=4,48l\)

d.\(n_{CuO}=\dfrac{m}{M}=\dfrac{32}{80}=0,4mol\)

\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)

0,4   >  0,2                                 ( mol )

 0,2       0,2            0,2                   ( mol )

\(m_{chất.rắn}=m_{CuO\left(dư\right)}+m_{Cu}=0,2.80+0,2.64=16+12,8=28,8g\)

\(\%m_{CuO}=\dfrac{16}{28,8}.100=55,55\%\)

\(\%m_{Cu}=100\%-55,55\%=44,45\%\)

\(n_{Zn}=\dfrac{16.25}{65}=0.25\left(mol\right)\)

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

\(0.25...................0.25.....0.25\)

\(V_{H_2}=0.25\cdot22.4=5.6\left(l\right)\)

\(m_{ZnCl_2}=0.25\cdot136=34\left(g\right)\)

\(n_{Zn}=\dfrac{3,25}{65}=0,05mol\)

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

   0,05    0,1          0,05       0,05

b)\(m_{ZnCl_2}=0,05\cdot136=6,8g\)

c)\(V_{H_2}=0,05\cdot22,4=1,12l\)

a, \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Zn + 2HCl -----> ZnCl₂ + H₂

0,2     0,4               0,2     0,2

b, \(m_{Zn}=0,2.65=13\left(g\right)\)

\(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

c, \(m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

\(a,n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl ---> ZnCl₂ + H₂

         0,2<--0,4<--------0,2<---0,2

\(b,\left\{{}\begin{matrix}m_{Zn}=0,1.65=13\left(g\right)\\m_{HCl}=0,4.36,5=14,6\left(g\right)\end{matrix}\right.\\ c,m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

a) Zn + 2HCl →ZnCl₂  + H₂

b) nZn = 6,5/65 = 0,1 mol . Theo tỉ lệ pư => nH₂ = nZn = nZnCl₂ =0,1 mol <=> V_H2(đktc) = 0,1.22,4 = 2,24 lít.

c) mZnCl₂ = 0,1 . 136 = 13,6 gam

d) nHCl =2nZn = 0,2 mol => mHCl = 0,2.36,5= 7,3 gam

Cách 2: áp dụng định luật BTKL => mHCl = mZnCl₂ + mH₂ - mZn 

<=> mHCl = 13,6 + 0,1.2 - 6,5 = 7,3 gam

a) PTHH: Zn + 2HCl ===> ZnCl₂ + H₂

b) Số mol kẽm: n_Zn = \(\frac{26}{65}=0,4\left(mol\right)\)

Theo phương trình, n_ZnCl2 = n_Zn = 0,4 (mol)

=> Khối lượng ZnCl₂ thu được: m_ZnCl2 = 0,4 x 136 = 54,4 gam

a) PTHH: Zn +2HCl -> ZnCl₂ + H₂\(\uparrow\)

Theo đề bài, ta có:

n_Zn=\(\frac{m_{Zn}}{M_{Zn}}=\frac{26}{65}=0,4\left(mol\right)\)

Theo PTHH và đề bài, ta có:

\(n_{Zncl_2}=n_{Zn}\)=0,4(mol)

Khối lượng muối kẽm clorua (ZnCl₂) tạo thành:

\(m_{ZnCl_2}=n_{ZnCl_2}.M_{ZnCl_2}=0,4.136=54,4\left(g\right)\)

câu C mô rồi

\(PTHH:Zn+2HCl->ZnCl_2+H_2\)

ap dung DLBTKL ta co

\(m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\)

\(=>m_{H_2}=m_{Zn}+m_{HCl}-m_{ZnCl_2}\\ =>m_{H_2}=13+14,6-27,2\\ =>m_{H_2}=0,4\left(g\right)\)

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

b, Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

c, \(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,25}{1}\), ta được CuO dư.

Theo PT: \(n_{CuO\left(pư\right)}=n_{H_2}=0,2\left(mol\right)\Rightarrow n_{CuO\left(dư\right)}=0,25-0,2=0,05\left(mol\right)\)

\(\Rightarrow m_{CuO\left(dư\right)}=0,05.80=4\left(g\right)\)