a) \(n_{H_2}=\dfrac{1,2.10^{23}}{6.10^{23}}=0,2\left(mol\right)\); \(n_{SO_2}=\dfrac{6,4}{64}=0,1\left(mol\right)\)

V = (1,5 + 2,5 + 0,2 + 0,1).22,4 = 96,32 (l)

b) \(m_{hh}=1,5.32+2,5.28+0,2.2+6,4=124,8\left(g\right)\)

\(n_{H_2}=\dfrac{1,2.10^{23}}{6.10^{23}}=0,2\left(mol\right)\)

\(n_{SO_2}=\dfrac{6,4}{64}=0,1\left(mol\right)\)

=> V_hh = (1,5 + 2,5+ 0,2 +0,1).22,4 = 96,32(l)

m_hh = 1,5.32 + 2,5.28 + 0,2.2 + 6,4 = 124,8(g)

Gọi số mol H₂, O₂ là a, b (mol)

=> \(\left\{{}\begin{matrix}a+b=\dfrac{22,4}{22,4}=1\\M_B=\dfrac{2a+32b}{a+b}=5,5.2=11\left(g/mol\right)\end{matrix}\right.\)

=> a = 0,7 (mol); b = 0,3 (mol)

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

Xét tỉ lệ: \(\dfrac{0,7}{2}>\dfrac{0,3}{1}\) => H₂ dư, O₂ hết

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

            0,6<--0,3------->0,6

=> \(\left\{{}\begin{matrix}m_{H_2O}=0,6.18=10,8\left(g\right)\\m_{H_2\left(dư\right)}=\left(0,7-0,6\right).2=0,2\left(g\right)\end{matrix}\right.\)

nX = 0,672/22,4 = 0,03 (mol)

Gọi nN2 = a (mol); nO2 = b (mol)

a + b = 0,03

28a + 32b = 0,88 

=> a = 0,02 (mol); b = 0,01 (mol)

%VN2 = 0,02/0,03 = 66,66%

%VO2 = 100% - 66,66% = 33,34%

M(X) = 0,88/0,03 = 88/3 (g/mol)

nX = 2,2 : 88/3 = 0,075 (mol)

VH2 = VX = 0,075 . 22,4 = 1,68 (l)

nhh = 8.96/22.4 = 0.4 (mol) 

nH2 = nO2 = 0.4/2 = 0.2 (mol) 

Đặt : nO2 (pư) = x (mol) 

      2H2 +   O2 -to-> 2H2O 

Bđ: 0.2____0.2

Pư: 2x_____x______2x

Kt : 0.2-2x__0.2-2x__2x 

M = 10*2 = 20 (g/mol) 

=> (0.2-2x)*2 + (0.2-2x)*32 + 2x * 18 = (0.2-2x+0.2-2x+2x)*20 

=> x = 0.06

H% = 0.06*2/0.2 * 100% = 60%

\(n_{hh_1}=\dfrac{V_{hh_1}}{22,4}=0,4\left(mol\right)\)

\(\Rightarrow n_{H_2}=n_{O_2}=0,2\left(mol\right)\)

PTHH: \(2H_2+O_2\rightarrow2H_2O\)

.............2............1................2.........

Sản phẩm sau khi nổ gồm \(H_2O\left(n=2x\right);H_2\left(\text{dư};n=0,2-2x\right);O_2\left(\text{dư};n=0,2-x\right)\)

Mà hỗn hợp khí và hơi có tỉ khối so với H2 là 10 nên:

\(d_{\dfrac{hh_2}{H_2}}=\dfrac{M_{hh_2}}{2}=10\\ \Rightarrow M_{hh_2}=20\left(g/mol\right)\)

Ta có pt: \(18\cdot2x+2\left(0,2-2x\right)+32\left(0,2-x\right)=20\left(2x+0,2-2x+0,2-x\right)\)

Giải pt ta được \(x=0,06\)

\(H=\dfrac{2x}{0,2}\cdot100\%=\dfrac{2\cdot0,06}{0,2}\cdot100\%=60\%\)

\(n_{SO_2}=\dfrac{6,4}{64}=0,1mol\)

\(n_{H_2}=\dfrac{1,2.10^{23}}{6.10^{23}}=0,2mol\)

\(\Rightarrow V_{hh}=\left(0,1+0,2+1,5+2,5\right).22,4=96,32l\)

\(m_{O_2}=1,5.32=48g\)

\(m_{N_2}=2,5.28=70g\)

\(m_{H_2}=0,2.2=0,4g\)

=> \(m_{hh}=48+70+0,4+6,4==124,8g\)

a)

\(V_{H_2} = V_{O_2} = \dfrac{4,48}{2} = 2,24(lít)\)

b)

\(2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ V_{H_2} = 2,24 < 2V_{O_2} = 4,48(lít)\)

Do đó, O₂ dư.

\(V_{O_2\ pư} = \dfrac{1}{2}V_{H_2} = 1,12(lít)\\ \Rightarrow V_A = V_{O_2\ dư} = 2,24 - 1,12 = 1,12(lít)\)

a, Theo bài ra ta có : \(\left\{{}\begin{matrix}V_{H2}+V_{O2}=4,48\\V_{H2}=V_{O2}\end{matrix}\right.\)

\(\Rightarrow V_{O2}=V_{H2}=2,24\left(l\right)\)

b, Ta có : \(n_{O_2}=n_{H_2}=\dfrac{V}{22,4}=0,1\left(mol\right)\)

\(PTHH:2H_2+O_2\rightarrow2H_2O\)

Thấy sau phản ứng O2 dư .

=> \(V_A=V_{O2du}=22,4\left(n_{o2}-n_{O2pu}\right)=1,12\left(l\right)\)

Vậy ..