PT: \(2Pb\left(NO_3\right)_2\underrightarrow{t^o}2PbO+4NO_2+O_2\)

Ta có: \(n_{Pb\left(NO_3\right)_2}=\dfrac{66,2}{331}=0,2\left(mol\right)\)

_ Nếu Pb(NO₃)₂ pư hết thì n_PbO = n_Pb(NO3)2 = 0,2 (mol)

⇒ m_PbO = 0,2.223 = 44,6 (g) < m _{chất rắn.}

⇒ Pb(NO₃)₂ còn dư.

Giả sử: \(n_{Pb\left(NO_3\right)_2\left(pư\right)}=x\left(mol\right)\)

⇒ n _{Pb(NO3)2 dư} = 0,2 - x (mol)

Theo PT: \(n_{PbO}=n_{Pb\left(NO_3\right)_2\left(pư\right)}=x\left(mol\right)\)

Có: m _{chất rắn} = m_PbO + m_{Pb(NO3)2 dư}

⇒ 55,4 = 223x + (0,2 - x).331

⇒ x = 0,1 (mol)

Theo PT: \(\left\{{}\begin{matrix}n_{NO_2}=2n_{Pb\left(NO_3\right)_2\left(pư\right)}=0,2\left(mol\right)\\n_{O_2}=\dfrac{1}{2}n_{Pb\left(NO_3\right)_2}=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow V_{khi}=V_{NO_2}+V_{O_2}=0,2.22,4+0,05.22,4=5,6\left(l\right)\)

Bạn tham khảo nhé!

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: 

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

\(a,n_{CuO}=\dfrac{59}{80}=0,7375\left(mol\right)\\ PTHH:2Cu\left(NO_3\right)_2\rightarrow^{t^o}2CuO+4NO_2\uparrow+O_2\uparrow\\ \Rightarrow\left\{{}\begin{matrix}n_{O_2}=\dfrac{1}{2}n_{CuO}=0,36875\left(mol\right)\\n_{NO_2}=2n_{CuO}=1,475\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,36875\cdot22,4=8,26\left(l\right)\\V_{NO_2}=1,475\cdot22,4=33,04\left(l\right)\end{matrix}\right.\)

\(b,\text{Chất rắn thu đc là }CuO\text{ gồm có }Cu,O\\ \%_O=\dfrac{16}{80}\cdot100\%=20\%\\ \Rightarrow m_O=59\cdot20\%=11,8\left(g\right)\\ \Rightarrow m_{Cu}=59-11,8=47,2\left(g\right)\)

ủa anh minh lm r mà trong này nè

Tham khảo:

https://hoc24.vn/cau-hoi/nung-752-gam-cuno32-bi-phan-huy-theo-so-do-phan-ung-sau-cuno32-cuo-no2-o2-sau-mot-thoi-gian-thay-con-lai-59-gam-chat-ran-a-tinh-the-ti.3307073058847

\(n_{Cu\left(NO_3\right)_2}=\dfrac{75,2}{188}=0,4mol\)

Gọi \(n_{Cu\left(NO_3\right)_2pứ}=x\left(mol\right)\)

\(Cu\left(NO_3\right)_2\underrightarrow{t^o}CuO+2NO_2+O_2\)

\(m_{CuO}=80x\left(g\right)\)

\(m_{Cu\left(NO_3\right)_2pứ}=188\cdot\left(0,4-x\right)mol\)

\(\Rightarrow m_{CuO}+m_{Cu\left(NO_3\right)_2pứ}=59\)

\(\Rightarrow x=0,15mol\)

\(V_{NO_2}=2\cdot0,15\cdot22,4=6,72l\)

\(V_{O_2}=0,15\cdot22,4=3,36l\)

\(m_{CuO}=0,15\cdot80=12g\)

Đặt : 

\(n_{O_2}=a\left(mol\right)\)

\(2Cu\left(NO_3\right)_2\underrightarrow{^{^{t^0}}}2CuO+4NO_2+O_2\)

\(2a..............2a.......4a......a\)

\(BTKL:\)

\(m_{khí}=75.2-59=16.2\left(g\right)\)

\(\Rightarrow4a\cdot46+32a=16.2\)

\(\Rightarrow a=0.075\)

\(V_{NO_2}=0.075\cdot4\cdot22.4=6.72\left(l\right)\)

\(V_{O_2}=0.075\cdot22.4=1.68\left(l\right)\)

Chất rắn gồm : Cu(NO₃)₂ dư , CuO 

\(m_{CuO}=0.075\cdot2\cdot80=12\left(g\right)\)

\(m_{Cu\left(NO_3\right)_2\left(dư\right)}=75.2-0.075\cdot2\cdot188=47\left(g\right)\)

Ta có

n Pb(NO3)2=66,2/331=0,2(mol)

Theo pthh

n PhO=n Pb(NO3)2=0,2(mol)

m PbO=0,2.223=44,6(g)

H=\(\frac{44,6}{55,4}.100\%=80,5\%\)

n_Pb(NO3)2 = \(\dfrac{66,2}{331}\) = 0,2 mol

do Pb(NO₃)₂ không phản ứng hết (dư)

=> gọi n_{Pb(NO3)2 pứ} = x mol

Pb(NO₃)₂ \(^{to}\rightarrow\) PbO + 2NO₂ \(\uparrow\)+ \(\dfrac{1}{2}\) O₂ \(\uparrow\)

x------------------------->2x -----> 0,5x

- ta có

m_NO2 + m_O2 = khối lượng chất rắn giảm

\(\Leftrightarrow\)2x . 46 + 0,5x . 32 = 66,2 - 55,4 = 10,8

=> x = 0,1 mol

=> V_NO2 = 0,2 . 22,4 = 4,48 (l)

=>V_O2 = 0,05 . 22,4 = 1,12 (l)

d_NO2/kk = \(\dfrac{46}{29}\)

d_O2/kk = \(\dfrac{32}{29}\)

2Pb(NO₃)₂ \(^{to}\rightarrow\) 2PbO + 4NO₂ \(\uparrow\)+ O₂ \(\uparrow\)

n_Pb(NO3)2 = \(\dfrac{66,2}{331}\) = 0,2 mol

m_khí = 66,2 - 55,4 = 10,8 g

- gọi số mol của NO₂ , O₂ lần lượt là x , y

- ta có :

46x + 32y = 10,8

\(\Leftrightarrow\) x = 4y

\(\Leftrightarrow\left\{{}\begin{matrix}x=0,2mol\\y=0,05mol\end{matrix}\right.\)

V_NO2 = 0,2 . 22,4 = 4,48(l)

V_O2 = 0,05 . 22,4 = 1,12(l)

D\(\dfrac{hh}{29}\) = \(\dfrac{46+32}{29}\) = 2,68 lần

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

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

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

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

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