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![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,n_{CuO}=\dfrac{59}{80}=0,7375\left(mol\right)\\ PTHH:2Cu\left(NO_3\right)_2\rightarrow^{t^o}2CuO+4NO_2\uparrow+O_2\uparrow\\ \Rightarrow\left\{{}\begin{matrix}n_{O_2}=\dfrac{1}{2}n_{CuO}=0,36875\left(mol\right)\\n_{NO_2}=2n_{CuO}=1,475\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,36875\cdot22,4=8,26\left(l\right)\\V_{NO_2}=1,475\cdot22,4=33,04\left(l\right)\end{matrix}\right.\)
\(b,\text{Chất rắn thu đc là }CuO\text{ gồm có }Cu,O\\ \%_O=\dfrac{16}{80}\cdot100\%=20\%\\ \Rightarrow m_O=59\cdot20\%=11,8\left(g\right)\\ \Rightarrow m_{Cu}=59-11,8=47,2\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(n_{Cu\left(NO_3\right)_2pứ}=x\left(mol\right)\)
\(Cu\left(NO_3\right)_2\underrightarrow{t^o}CuO+2NO_2+O_2\)
\(m_{CuO}=80x\left(g\right)\)
\(m_{Cu\left(NO_3\right)_2pứ}=188\cdot\left(0,4-x\right)mol\)
\(\Rightarrow m_{CuO}+m_{Cu\left(NO_3\right)_2pứ}=59\)
\(\Rightarrow x=0,15mol\)
\(V_{NO_2}=2\cdot0,15\cdot22,4=6,72l\)
\(V_{O_2}=0,15\cdot22,4=3,36l\)
\(m_{CuO}=0,15\cdot80=12g\)
![](https://rs.olm.vn/images/avt/0.png?1311)
nCu(NO3)2=15.04188=0.08(mol)
- Gọi số mol Cu(NO3)2 đã phản ứng là x.
⇒⇒mCuO=80.x (g)
và mCu(NO3)2phản ứng =188.(0.08-x) (g)
- Theo bài ra, ta có:
mCuO+mCu(NO3)2=8.56(g)
⇒⇒80.x + 188.(0.08-x) = 8.56
⇒⇒108x = 6.48
⇒⇒x=0.06 (mol)
⇒mCu(NO3)2pư=0.06⋅188=11.28(g)
⇒%mCu(NO3)2pư=11.28\15.04⋅100%=75%
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Cu\left(NO_3\right)_2}=\dfrac{15,04}{188}=0,08\left(mol\right)\)
Gọi số mol Cu(NO3)2 pư là a
PTHH: 2Cu(NO3)2 --to--> 2CuO + 4NO2 + O2
a---------------->a------->2a
=> 188(0,08-a) + 80a = 8,56
=> a = 0,06
=> \(\left\{{}\begin{matrix}m_{Cu\left(NO_3\right)_2}=\left(0,08-0,06\right).188=3,76\left(g\right)\\m_{CuO}=0,06.80=4,8\left(g\right)\end{matrix}\right.\)
=> mCu(NO3)2 bị phân hủy = 0,06.188 = 11,28(g)
Có lẽ bài này nhường chỗ cho idol của mình là Quang Nhân :))
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Cu\left(NO_3\right)_2}=\dfrac{15.04}{188}=0.08\left(mol\right)\)
\(n_{O_2}=x\left(mol\right)\)
\(2Cu\left(NO_3\right)_2\underrightarrow{t^0}2CuO+4NO_2+O_2\)
\(2x......................4x......x\)
\(BTKL:\)
\(m_{NO_2}+m_{O_2}=15.04-8.56=6.48\left(g\right)\)
\(\Rightarrow4x\cdot46+32x=6.48\)
\(\Rightarrow x=0.03\)
\(\%Cu\left(NO_3\right)_{2\left(ph\right)}=\dfrac{0.03}{0.08}\cdot100\%=37.5\%\)
\(b.\)
\(\overline{M}=\dfrac{6.48}{0.12+0.03}=43.2\left(\dfrac{g}{mol}\right)\)
\(d_{\dfrac{hh}{H_2}}=\dfrac{43.2}{2}=21.6\)
\(c.\)
\(H\%=\dfrac{0.03}{0.08}\cdot100\%=37.5\%\)
\(\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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?
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi \(n_{Cu\left(NO3\right)2\left(pư\right)}:a\left(mol\right)\)
\(n_{Cu\left(NO3\right)2\left(bđ\right)}=0,08\left(mol\right)\)
\(PTHH:2Cu\left(NO_3\right)_2\rightarrow2CuO+O_2+4NO_2\)
_____________a_______a__________0,5a____2a
\(n_{Cu\left(NO3\right)2\left(dư\right)}=0,08-a\left(mol\right)\)
\(n_{CuO}=a\left(mol\right)\Rightarrow\left(0,08-a\right).188+80a=8,56\)
\(\Leftrightarrow a=0,06\left(mol\right)\)
\(\Rightarrow H=\frac{0,06}{0,08}.100\%=75\%\)
\(\left\{{}\begin{matrix}n_{NO2}=0,06.2=0,12\left(mol\right)\\n_{O2}=0,5a=0,03\left(mol\right)\end{matrix}\right.\)
\(M_{khi}=43,2\left(\frac{g}{mol}\right)\Rightarrow d_{M_{khi}/H2}=\frac{43,2}{2}=21,6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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:
\(H = \dfrac{0,1.331}{66,2}.100\% = 50\%\)
\(n_{Cu\left(NO_3\right)_2}=\dfrac{15.04}{188}=0.08\left(mol\right)\)
- Gọi số mol \(Cu\left(NO_3\right)_2\) đã phản ứng là x.
\(\Rightarrow\)\(m_{CuO}=80.x\) (g)
và \(m_{Cu\left(NO_3\right)_2}\)phản ứng =188.(0.08-x) (g)
- Theo bài ra, ta có:
\(m_{CuO}+m_{Cu\left(NO_3\right)_2}=8.56\left(g\right)\)
\(\Rightarrow\)80.x + 188.(0.08-x) = 8.56
\(\Rightarrow\)80.x + 15.04 - 188x = 8.56
\(\Rightarrow\)108x = 6.48
\(\Rightarrow\)x=0.06 (mol)
\(\Rightarrow m_{Cu\left(NO_3\right)_2}pư=0.06\cdot188=11.28\left(g\right)\)
\(\Rightarrow\%m_{Cu\left(NO_3\right)_2}pư=\dfrac{11.28}{15.04}\cdot100\%=75\%\)
PTHH:
2Cu(NO\(_3\))\(_2\) \(\rightarrow\) 2CuO + 4NO\(_2\)+ O\(_2\)
Mol: 0, 08 \(\rightarrow\) 0,08 : 0,16 : 0,04
Ta có: m\(_{Cu\left(NO_3\right)_2}\)= 15,04(g)
=> n\(_{Cu\left(NO_3\right)_2}\)= 15,04 : 188= 0,08 (mol)
Giả sử chất rắn chỉ có CuO
m\(_{rắn}\)= m\(_{CuO}\)= 0,08. 80 = 6,4(g) \(\ne\) 8,56(g)
Vậy trong chất rắn có CuO và Cu(NO\(_3\))\(_2\) dư
PTHH:
2Cu(NO\(_3\))\(_2\) \(\rightarrow\)2CuO + 4NO\(_2\)+ O\(_2\)
Mol: x \(\rightarrow\) x : 2x : 0,5x
Gọi x là số mol của Cu(NO\(_3\))\(_2\)
=> n\(_{Cu\left(NO_3\right)_{2_{dư}}}\)= 0,08 - x(mol)
=> m\(_{rắn}\)= m\(_{Cu\left(NO_3\right)_{2_{dư}}}\)+ m\(_{CuO}\)
<=> 8,56 = 188( 0,08 - x) + 80x
<=> 8,56 = 15,04 - 188x + 80x
<=> 188x - 80x = 15,04 - 8,56
<=> 108x = 6,48
<=> x = 0,06
m\(_{Cu\left(NO_3\right)_{2_{Pứ}}}\)= 0,06. 188= 11,28(g)
%m\(_{Cu\left(NO_3\right)_{2_{Pư}}}\) = \(\frac{11,28}{15,04}\). 100% = 75%