tham khảo 

Câu 1:

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{Zn}=\dfrac{16,25}{65}=0,25\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Zn}=0,25\left(mol\right)\Rightarrow V_{H_2}=0,25.22,4=5,6\left(l\right)\)

c, \(n_{HCl}=2n_{Zn}=0,5\left(mol\right)\Rightarrow m_{Zn}=0,5.36,5=18,25\left(g\right)\)

Câu 2:

a, \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

b, \(n_P=\dfrac{1,24}{31}=0,04\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{5}{4}n_P=0,05\left(mol\right)\Rightarrow V_{O_2}=0,05.22,4=1,12\left(l\right)\)

c, \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,02\left(mol\right)\Rightarrow m_{P_2O_5}=0,02.142=2,84\left(g\right)\)

1.

`Zn+2HCl->ZnCl_2+H_2`

0,25---0,5---------------0,25

`n_Zn=(16,25)/65=0,25 mol`

`=>V_(H_2)=0,25.22,4=5,6l`

`=>m_(HCl)=0,5.36,5=18,25g`

2.

`4P+5O_2->2P_2O_5`(to)

0,04---0,05----0,02

`nP=(1,24)\31=0,04 mol`

`V_(O_2)=0,05.22,4=1,12l`

`m_(P_2O_5)=0,02.142=2,84g`

a) Zn + 2HCl --> ZnCl₂ + H₂

b) \(n_{H_2}=\dfrac{22,4}{22,4}=1\left(mol\right)=>m_{H_2}=1.2=2\left(g\right)\)

Theo ĐLBTKL:

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

=> \(m_{HCl}=136+2-65=73\left(g\right)\)

\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ b,m_{H_2}=\dfrac{22,4}{22,4}.2=2(g)\\ BTKL:m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\\ \Rightarrow m_{HCl}=136+2-65=73(g)\)

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

ap dung DLBTKL ta co

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

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

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

Khối lượng của `HCl` đã phản ứng:

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

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

a) Zn+2HCl-->ZnCl2+H2

b) Theo định luật bảo toàn khối lượng ta có công thức sau

m\(_{ZnCl2}+m_{H2}=m_{Zn}+m_{HCl}\)

c) Ta có

m\(_{H2}=m_{Zn}+m_{HCl}-m_{ZnCl2}\)

=\(13-14,6-27,2=0,4\left(g\right)\)

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

b) CT về khối lượng: \(m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\)

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

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

Hiện tượng: viên kẽm tan dần, có khí không màu thoát ra.

\(b,n_{Zn}=\dfrac{9,75}{65}=0,15\left(mol\right)\\ \Rightarrow n_{HCl}=2n_{Zn}=0,3\left(mol\right)\\ \Rightarrow m_{HCl}=0,3\cdot36,5=5,475\left(g\right)\\ c,n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,15\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,15\cdot136=20,4\left(g\right)\\m_{H_2}=0,15\cdot2=0,3\left(g\right)\end{matrix}\right.\\ \Rightarrow m_{dd_{ZnCl_2}}=9,75+100-0,3=109,45\left(g\right)\\ \Rightarrow C\%_{dd_{ZnCl_2}}=\dfrac{20,4}{109,45}\cdot100\%\approx18,64\%\)

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

\(Zn+2HCl\xrightarrow[]{}ZnCl_2+H_2\uparrow\)

0,1         →          0,1 → 0,1

a) \(V_{H_2}=22,4\cdot0,1=2,24\left(l\right)\)

b) \(m_{ZnCl_2}=0,1\cdot136=13,6\left(g\right)\)

c) \(Zn+H_2SO_4\xrightarrow[]{}ZnSO_4+H_2\uparrow\)

bđ: 0,1 → 0,4

pư: 0,1 → 0,1

\(\Rightarrow H_2SO_4\text{ dư}\)

\(\Rightarrow n_{H_2SO_4\text{ dư}}=0,4-0,1=0,3\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4\text{ dư}}=0,3\cdot98=29,4\left(g\right)\)

bài này quá dễ nha.Em dùng bảo toàn khối lượng là ra ngay nà

mZn + mHCl = mZnCl2 + mH2

=>mHCl =(mZnCl2 + mH2) - mZn

               = (340+5) -162.5= 182.5g

vậy khối lượng HCl phản ứng là 182.5g

Chúc em học tốt !!!!:))