Ta có: \(n_{ZnO}=\dfrac{16,2}{81}=0,2\left(mol\right)\)

a, PT: \(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

______0,2_____0,4____0,2 (mol)

b, \(m_{HCl}=0,4.36,5=14,6\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{14,6}{10\%}=146\left(g\right)\)

c, \(C\%_{ZnCl_2}=\dfrac{0,2.136}{16,2+146}.100\%\approx16,77\%\)

Bạn tham khảo nhé!

\(a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Áp.dụng.ĐLBTKL,ta.có:\\ m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\\ m_{H_2}=m_{Zn}+m_{HCl}-m_{ZnCl_2}=6,5+7,3-13,6=0,2\left(g\right)\)

Zn + 2HCl --> ZnCl₂ + H₂

\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)

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

\(V_{H_2}=0,4.22,4=8,96\left(l\right)\)

\(nZn=\dfrac{26}{65}=0,4mol\)

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

0,4-->0,8----->0,4------->0,4

\(mZnCl_2=136.0,4=54,4g\)

\(VH_2=0,4.22,4=8,96lít\)

a) Lập công thức hóa học của muối kẽm clorua:  Z n C l 2

b) Khối lượng muối Z n C l 2  = 6,5 + 7,3 – 0,2 = 13,6 (gam)

a) Zn + 2HCl --> ZnCl₂ + H₂

b) \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

            0,3------------->0,3--->0,3

=> m_ZnCl2 = 0,3.136 = 40,8 (g)

c) V_H2 = 0,3.22,4 = 6,72 (l)

a. \(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

b. \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

Mol theo PTHH : \(1:2:1:1\)

Mol theo phản ứng : \(0,3\rightarrow0,6\rightarrow0,3\rightarrow0,3\)

\(\Rightarrow m_{ZnCl_2}=n_{ZnCl_2}.M_{ZnCl_2}=0,3.\left(65+71\right)=40,8\left(g\right)\)

c. Từ b. \(\Rightarrow n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow V_{H_2}=n_{H_2}.22,4=0,3.22,4=6,72\left(l\right)\)

\(a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(b,n_{Zn}=\dfrac{m}{M}=\dfrac{16,25}{65}=0,25\left(mol\right)\\ Theo.PTHH:n_{HCl}=2.n_{Zn}=2.0,25=0,5\left(mol\right)\\ m_{HCl}=n.M=0,5.36,5=18,25\left(g\right)\)

\(Theo.PTHH:n_{H_2}=n_{Zn}=0,25\left(mol\right)\\ V_{H_2\left(đktc\right)}=n.22,4=0,25.22,4=5,6\left(l\right)\)

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

b)Khối lượng Zn:\(m_{Zn}=\dfrac{16,25}{65}=0,25\left(mol\right)\)

Ta có: \(n_{HCl}=2n_{Zn}=0,5\left(mol\right)\)

Khối lượng axit HCl cần dùng là: \(m_{HCl}=0,5.36,5=18,25\left(g\right)\)

c)Theo pt ta có: \(n_{H_2}=n_{Zn}=0,25\left(mol\right)\)

Thể tích H₂ là: \(V_{H_2}=n.22,4=0,25.22,4=5,6\left(ml\right)\)

\(n_{HCl}=0,25.2=0,5\left(mol\right)\\ a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{Zn}=n_{H_2}=n_{ZnCl_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ m_{Zn}=0,25.65=16,25\left(g\right)\\ c,V_{H_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\)

Không biết đúng không nữa;-;;;

a) PTHH: Zn + 2HCl -> ZnCl2 + H2 

b) HCl=250ml=0,25l

n2HCl= V/22,4= 0,5/22,4= 0,02(mol)

Zn  +  2HCl -> ZnCl2  +  H2

1          2              1         1

0,01 <-0,5--------------> 0,01

mZn= n.M= 0,01.65= 0,65(gam)

c) VH2=n . 22,4= 0,01 . 22,4= 0,224(l)
 

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

Theo PT: \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\Rightarrow V_{HCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)

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

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\\ a.Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

\(b)V_{ddHCl}=\dfrac{0,4}{2}=0,2l\\ m_{ZnCl_2}=0,2.136=27,2g\)

a) 

Zn  +  2HCl  →  ZnCl₂   +  H₂ 

b) nZn = \(\dfrac{3,5}{65}\)=\(\dfrac{7}{130}\) mol 

Theo tỉ lệ phản ứng => nH₂ = nZn= \(\dfrac{7}{130}\)mol

<=> V H₂ = \(\dfrac{7}{130}\).22,4 = 1,206 lít

c) nZnCl₂ = nZn => mZnCl₂ = \(\dfrac{7}{130}\).136= 7,32 gam