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

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

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 

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

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

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?

\(Cu\left(NO_3\right)_2\rightarrow CuO+2NO_2+\dfrac{1}{2}O_2\)

188 88 \(\rightarrow\) giảm 108(g)\(\Leftrightarrow1\left(mol\right)\)

giảm 54 \(\Leftrightarrow a\left(mol\right)\)

\(\Leftrightarrow108a=54\Leftrightarrow a=0,5\)

\(n_{Cu\left(NO_3\right)_2}=n_{CuO}=0,5\left(mol\right)\)

\(\rightarrow m_{Cu\left(NO_3\right)_2}=0,5.188=94\left(g\right)\)

\(n_{NO_2}=2n_{CuO}=1\left(mol\right);n_{O_2}=\dfrac{1}{2}n_{CuO}=0,25\left(mol\right)\)

\(\rightarrow V_k=22,4.\left(1+0,25\right)=28\left(l\right)\)

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%

Ko cần nữa mik giải đc r

pư thu là gì bn???

Câu 2:

\(n_X=\dfrac{1,5.10^{23}}{6.10^{23}}=0,25\left(mol\right)=>M_X=\dfrac{7}{0,25}=28\left(g/mol\right)\)

\(d_{X/H_2}=\dfrac{28}{2}=14\)

\(d_{X/CH_4}=\dfrac{28}{16}=1,75\)

\(d_{X/O_2}=\dfrac{28}{32}=0,875\)

X có thể là CO, C₂H₄, N₂

Câu 3:

a) \(n_{Cu\left(NO_3\right)_2}=\dfrac{282}{188}=1,5\left(mol\right)\)

=> \(n_{Cu\left(NO_3\right)_2\left(pư\right)}=\dfrac{1,5.90}{100}=1,35\left(mol\right)\)

PTHH: 2Cu(NO₃)₂ --t^o--> 2CuO + 4NO₂ + O₂

_______1,35------------->1,35------------>0,675

=> m_CuO = 1,35.80= 108(g)

=> V_O2 = 0,675.22,4 = 15,12(l)

b) Gọi số mol Cu(NO₃)₂ cần nung là a

=> \(n_{Cu\left(NO_3\right)_2\left(pư\right)}=\dfrac{a.90}{100}=0,9a\left(mol\right)\)

PTHH: 2Cu(NO₃)₂ --t^o--> 2CuO + 4NO₂ + O₂

________0,9a-------------------->1,8a--->0,45a

=> V_khí = (1,8a+0,45a).22,4 = 5

=> a = 0,0992 (mol)

=> m_Cu(NO3)2 = 0,0992.188 = 18,6496(g)

Đáp án A

Đáp án A.

Đáp án A