Sửa đề : 13,17 → 13,16

Gọi \(n_{Cu(NO_3)_2\ pư} = a(mol)\)

\(2Cu(NO_3)_2 \xrightarrow{t^o} 2CuO + 4NO_2 + O_2\\ n_{CuO} = n_{Cu(NO_3)_2\ pư} = a(mol)\\ \Rightarrow m_{chất\ rắn\ sau\ pư} = m_{CuO} + m_{Cu(NO_3)_2\ dư}\\ \Rightarrow 8,36 = 80a + 13,16 - 188a\\ \Rightarrow a = \dfrac{2}{45}\\ H = \dfrac{\dfrac{2}{45}.188}{13,16}.100\% = 63,49\%\)

sao sửa đề vậy anh..

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

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

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 \(n_{O_2}=x\left(mol\right)\)

\(2Cu\left(NO_3\right)_2\xrightarrow[]{t^o}2CuO+4NO_2\uparrow+O_2\\ \Rightarrow n_{NO_2}=4a\\ \Rightarrow22,4.\left(4a+a\right)=5,6\\ \Rightarrow a=0,05\left(mol\right)\\ \Rightarrow n_{Cu\left(NO_3\right)_2}=2a=0,1\left(mol\right)\\ \Rightarrow m_{Cu\left(NO_3\right)_2}=0,1.188=18,8\left(g\right)\)

0,05 tính sau v bn

Xin hỏi bn là 3,36 từ đâu mà ra

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?

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%

Cần câu b