Đặt \(n_{O_2}=x\left(mol\right)\)

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

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 (mol)

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

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

______0,9a---------------------->1,8a--->0,45a

=> (1,8a+0,45a).22,4 = 5

=> a = 0,0992 (mol)

=> \(m_{Cu\left(NO_3\right)_2}=0,0992.188=18,6496\left(g\right)\)

a. Ta có: n_CuO=1,5.2:2=1,5 (mol)

m_CuO=n.M=1,5.80=120 (g)

n_O2=1,5.1:2=0,75 (mol)

V_O2=n.22,4=0,75.22,4=16,8 (lit)

b.

pư thu là gì bn???

Chọn B

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

Đặt số mol O₂ là a à Số mol NO₂ là 4a; số mol Mg(NO₃)₂ = 2a

Theo bài ra ta có: 5a = 0,25 à a = 0,05 à số mol Mg(NO₃)₂ =  0,1

à Khối lượng Mg(NO₃)₂ = 14,8 gam

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 

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

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