Gọi \(n_{Pb(NO_3)_2\ pư} = a(mol)\)

2Pb(NO₃)₂   \(\xrightarrow{t^o}\)   2PbO  +   4NO₂   +   O₂

a............................................2a............0,5a.............(mol)

Ta có : 

\(m_{NO_2} + m_{O_2} = 66,2-55,4\\ \Rightarrow 2a.46 + 32.0,5a= 10,8\\ \Rightarrow a = 0,1\)

Vậy hiệu suất phản ứng: 

\(H = \dfrac{0,1.331}{66,2}.100\% = 50\%\)

\(2Pb\left(NO3\right)2-->2PbO+4NO2+O2\)

\(n_{Pb\left(NO3\right)2}=\frac{33,1}{331}=0,1\left(mol\right)\)

\(n_{PbO}=n_{Pb\left(NO3\right)2}=0,1\left(mol\right)\)

\(m_{PbO}=0,1.223=22,3\left(g\right)\)

\(H\%=\frac{22,3}{27,7}.100\%=80,5\%\)

nPb(NO3)₂=0,2mol.
Gọi \(n_{Pb\left(NO_3\right)_2}\) bị nhiệt phân = x mol.
Pb(NO₃)₂ ---> PbO + 2NO₂ +\(\frac{1}{2}O_2\)
x------------------------------->2x--------->0,5x mol
Ta có: mNO₂+mO₂= khối lượng chất rắn giảm.
=> 2x.46+0,5x.32=66,2-55,4=10,8 => x=0,1
=> H=50%

Đáp án B

\(a,2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2\\ n_{KMnO_4}=\dfrac{395}{158}=2,5(mol)\\ \Rightarrow n_{O_2}=1,25(mol)\\ \Rightarrow V_{O_2}=1,25.22,4=28(l)\\ \Rightarrow V_{O_2(tt)}=28.85\%=23,8(l)\)

\(b,n_{O_2}=\dfrac{67,2}{22,4}=3(mol)\\ 2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2\\ \Rightarrow n_{KMnO_4}=6(mol)\\ \Rightarrow m_{KMnO_4}=6.158=948(g)\\ \Rightarrow m_{KMnO_4(tt)}=\dfrac{948}{80\%}=1185(g)\)

a) PTHH: \(2Cu\left(NO_3\right)_2\xrightarrow[]{t^o}2CuO+4NO_2+O_2\)

Gọi \(n_{O_2}=a\left(mol\right)\Rightarrow n_{NO_2}=4a\left(mol\right)\)

Bảo toàn khối lượng: \(m_{Cu\left(NO_3\right)_2}=m_{rắn}+m_{khí}\)

\(\Rightarrow m_{khí}=m_{Cu\left(NO_3\right)_2}-m_{rắn}=6,48\left(g\right)=32a+46\cdot4a\) \(\Rightarrow a=0,03\left(mol\right)\)

\(\Rightarrow n_{Cu\left(NO_3\right)_{21}\left(p.ứ\right)}=0,06\left(mol\right)\) \(\Rightarrow m_{Cu\left(NO_3\right)_2\left(p.ứ\right)}=0,06\cdot188=11,28\left(g\right)\)

b) Ta có: \(\overline{M}_{khí}=\dfrac{0,03\cdot32+0,03\cdot4\cdot46}{0,03+0,03\cdot4}=43,2\) \(\Rightarrow d_{khí/H_2}=\dfrac{43,2}{2}=21,6\)

c) Ta có: \(H\%=\dfrac{m_{Cu\left(NO_3\right)_2\left(p.ứ\right)}}{m_{Cu\left(NO_3\right)_2\left(bđ\right)}}=\dfrac{11,28}{15,04}=75\%\)

Đề không cho bất kì khối lượng hay con số nào sao em?

2Pb(NO3)2 --------> 2PbO+4NO2+O2

\(n_{Pb\left(NO_3\right)_2}=\frac{66,2}{331}=0,2mol\)

\(n_{PbO}=n_{Pb\left(NO_3\right)_2}=0,2mol\)

\(m_{PbO}=0,2.223=44,6g\)

\(H=\frac{44,6}{55,4}.100\%=80,505\%\)