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)

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

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 

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

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

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

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

Đáp án B

Đặt n_N2 = x; n_H2 = y ⇒ n_A = x + y = 0,05 mol; m_A = 28x + 2y = 0,05 × 11,4 × 2

Giải hệ có: x = 0,04 mol; y = 0,01 mol || n_O/khí = 2.∑n_{(NO2, O2)} . Bảo toàn nguyên tố Oxi: 

n_O/H2O = 0,25 × 6 - 0,45 × 2 = 0,6 mol. Bảo toàn nguyên tố Hidro: n_NH4⁺ = 0,02 mol.

n_Cl^–_/Z = n_HCl = 1,3 mol. Bảo toàn điện tích: n_Mg²⁺ = (1,3 - 0,25 × 2 - 0,02)/2 = 0,39 mol.

⇒ m = 0,39 × 24 + 0,25 × 64 + 0,02 × 18 + 1,3 × 35,5 = 71,87(g)

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?

1)

$M_X = 1,375.32  = 44(g/mol)$
$M_X = 0,0625.32 = 2(g/mol)$

2)

$M_X = 2,207.29 = 64(g/mol)$

$M_X = 1,172.29 = 34(g/mol)$

3)

$M_X = 17.2 = 34(g/mol)$

Vậy khí X là $H_2S$

4)

a) $M_X = 0,552.29 = 16$
Gọi CTHH của X là $C_xH_y$

Ta có :  $\dfrac{12x}{75} = \dfrac{y}{25} = \dfrac{16}{100}$

Suy ra:  x = 1 ; y = 4

Vậy X là $CH_4$

$CH_4 + 2O_2 \xrightarrow{t^o} CO_ 2+ 2H_2O$
$V_{O_2} = 2V_{CH_4} = 11,2.2 = 22,4(lít)$

Gọi số mol NO₂, O₂ là a, b

Có: \(\dfrac{46a+32b}{a+b}=19,5.2=39\)

=> 7a = 7b

=> a = b

PTHH: 2Mg(NO₃)₂ --t^o--> 2MgO + 4NO₂ + O₂

________0,5a<----------------------a----->0,25a

2NaNO₃ --t^o--> 2NaNO₂ + O₂

(2b-0,5a)<------------(b-0,25a)

=> m_NaNO3 = (2b-0,5a).85 = 127,5a (g)

=> m_Mg(NO3)2 = 0,5a.148 = 74a (g)

=> \(\left\{{}\begin{matrix}\%NaNO_3=\dfrac{127,5a}{127,5a+74a}.100\%=63,275\%\\\%Mg\left(NO_3\right)_2=\dfrac{74a}{127,5a+74a}.100\%=36,725\%\end{matrix}\right.\)

Hahaha

Câu1) n_CO2 =m/M=11/44=0,25(mol)

         nH₂= 9.10²³/6.10²³=1,5(mol)

     V_H2 =n.22,4=1,5.22,4=33,6(l)