Tính mol như thường thôi

Bài 1:

\(PTHH:2HgO\underrightarrow{Phân.hủy}2Hg+O_2\\ á,Theo.PTHH:n_{O_2}=\dfrac{1}{2}.n_{HgO}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\\ V_{O_2\left(đktc\right)}=n.22,4=0,05.22,4=1,12\left(l\right)\)

\(b,n_{HgO}=\dfrac{m}{M}=\dfrac{43,4}{217}=0,2\left(mol\right)\\ Theo.PTHH:n_{Hg}=n_{HgO}=0,2\left(mol\right)\\ m_{Hg}=n.M=0,2.201=40,2\left(g\right)\)

\(c,n_{Hg}=\dfrac{m}{M}=\dfrac{14,07}{201}=0,07\left(mol\right)\\ Theo.PTHH:n_{HgO}=n_{Hg}=0,07\left(mol\right)\\ m_{HgO}=n.M=0,07.217=15,19\left(g\right)\)

Câu 2:

\(a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Theo.PTHH:n_{Zn}=n_{H_2}=0,3\left(mol\right)\\ m_{Zn}=n.M=0,3.65=19,5\left(g\right)\\ b,Theo.PTHH:n_{HCl}=2.n_{Zn}=2.0,3=0,6\left(mol\right)\\ m_{HCl}=n.M=0,6.36,5=21,9\left(g\right)\)

b,  số mol HgO phân hủy là:

43,4/217 = 0,2 ( mol)

theo ptpư: nHgO= nHg = 0,2 (mol)

khối lượng thủy ngân sịnh ra khi cho 43,4g HgO phân hủy là:

mHg = 201*0,2=40,2 g

2HgO -t--> 2Hg + O2 
  0,15-------------->0,75 (mol) 
=> mO2 = 0,75 . 32 = 24 (g)

a, Ta có:

\(n_{HgO}=\frac{21,7}{217}=0,1\left(mol\right)\)

\(\Rightarrow n_{O2}=n_{HgO}=0,1\left(mol\right)\)

\(\Rightarrow V_{O2}=0,1.22,4=2,24\left(l\right)\)

b, Ta có:

\(n_{HgO}=0,2\left(mol\right)\Rightarrow n_{Hg}=n_{HgO}=0,2\left(mol\right)\)

\(\Rightarrow m_{Hg}=0,2.201=40,2\left(g\right)\)

c,Ta có:

\(n_{Hg}=0,07\left(mol\right)\Rightarrow n_{HgO}=0,07\left(mol\right)\)

\(\Rightarrow m_{HgO}=0,07.217=15,19\left(g\right)\)

bạn cân bằng PTHH rồi tính số mol các chất ra sau đó dựa vào PTHH mà tính số mol nhé

nHgO=2,17/217=0,01(mol)

nO2=0,112/22,4=0,005(mol)

PTHH: 2 Hg + O2 -to-> 2 HgO

Ta có: 0,01/2 = 0,005

=>P.ứ xảy ra hết.

=> nHg=nHgO=0,01(mol)

=>mHg=0,01.201=2,01(g)

\(n_{HgO}=\dfrac{2,17}{217}=0,01\left(mol\right)\)

\(n_{O_2}=\dfrac{0,112}{22,4}=0,005\left(mol\right)\)

PTHH :       \(2HgO\rightarrow2Hg+O_2\)

                   0,01    0,0025    0,005          (mol)

\(m_{Hg}=0,0025.201=0,5025\left(g\right)\)

2HgO---->2Hg+O2

a) n O2=1/2n HgO=4(mol)

m O2=4.32=128(g)

b) n HgO=434/217=2(mol)

Theo pthh

n Hg=n HgO=2(mol)

m Hg=2.201=402(g)

c) n Hg=150,75/201=0,75(mol)

Theo pthh

n HgO=n Hg=0,75(mol)

m HgO=0,75.217=162,75(g)

Ta có : \(n_{HgO}=\frac{54,25}{217}=0,25\left(mol\right)\)

PTHH : \(2HgO\rightarrow2Hg+O_2\uparrow\)

Theo phương trình, ta có : \(n_{O_2}=\frac{1}{2}n_{HgO}\Rightarrow n_{O_2}=0,125\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,125.22,4=2,8\left(l\right)\)

2. 
a) 2Na + O2 -> 2NaO

b) P2O5 + 3H2O -> 2H3PO4 

c) HgO -> Hg + 1/2O2 

d) 2Fe(OH)3 -> Fe2O3 + 3H2O 

e) Na2CO3 + CaCl2 -> CaCO3 + 2NaCl 

\(n_{KMnO4} = \dfrac{15,8}{158} = 0,1 (mol) \\ PTHH: 2KMnO_4 \rightarrow (t^o) K_2MnO_4 + MnO_2 + O_2 \\ Mol: 0,1 \rightarrow 0,05 \rightarrow 0,05 \rightarrow 0,05 \\ 3Fe + 2O_2 \rightarrow (t^o) Fe_3O_4 \\ Mol:0,075 \leftarrow0,05 \leftarrow 0,025 \\ m_{Fe_3O_4} = 232 . 0,025 = 5,8(g)\)