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Ta có :
\(m_x=m_{CO_2}+m_{O_2}=44.0,15+32.0,1=9,8\) gam
\(n_x=n_{CO_2}+n_{O_2}=0,15+0,1=0,25\) mol
\(\Rightarrow\overline{M_x}=\frac{m_x}{n_x}=\frac{9,8}{0,25}=39,2\) gam/mol
\(\Rightarrow d_{\frac{x}{kk}}=\frac{\overline{M_x}}{29}=\frac{39,2}{29}=1,35\)
\(a.\overline{M}_A=\dfrac{n_{O_2}\cdot M_{O_2}+n_{N_2}\cdot M_{N_2}+n_{CO_2}\cdot M_{CO_2}+n_{H_2}\cdot M_{H_2}}{n_{O_2}+n_{N_2}+n_{CO_2}+n_{H_2}}\\ =\dfrac{0,2\cdot32+0,1\cdot28+0,05\cdot44+0,15\cdot2}{0,2+0,1+0,05+0,15}\\ =\dfrac{11,7}{0,5}=23,4\left(g/mol\right)\)
b) \(d_{hh/CH_4}=\dfrac{23,4}{16}=1,4625\)
a/ Ta có: VO2(đktc) = 0,25 x 22,4 = 5,6 lít
b/ Ta có: VH2(đktc) = 0,6 x 22,4 = 13,44 lít
c/ Ta có:
- nCO2 = 4,4 / 44 = 0,1 (mol)
- nN2 = 22,8 / 28 \(\approx0,81\left(mol\right)\)
=> Vhỗn hợp khí(đktc) = ( 0,1 + 0,15 + 0,81 ) x 22,4 = 23,744 (lít)
a.VO2=n.22,4=0,25.22,4=5,6l
b.VH2=n.22,4=0,6.22,4=13,44l
c.nCO2=m:M=4,4:44=0,1mol
nN2=m:M=22,8:28=0,8mol
Vhh=(0,1.22,4)+(0,15.22,4)+(0,8.22,4)=23,52l
a)
Mg + H2SO4 --> MgSO4 + H2
2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
Fe + H2SO4 --> FeSO4 + H2
b)
PTHH: Mg + H2SO4 --> MgSO4 + H2
0,1-->0,1---------------->0,1
2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
0,04-->0,06----------------->0,06
Fe + H2SO4 --> FeSO4 + H2
0,15-->0,15------------->0,15
=> a = nH2SO4 = 0,1 + 0,06 + 0,15 = 0,31 (mol)
m = mX - mH2 = 0,1.24 + 0,04.27 + 0,15.56 - 2(0,1 + 0,06 + 0,15)
= 11,26 (g)
a) 0,02 mol Hiđrô
mH = nH.MH = 0,02.1= 0,02 (g)
VH = nH.22,4 = 0,02.22,4= 0,448 (l)
(Các câu sau làm tương tự bạn nhé) 
a. \(m_{H_2}=n.M=0,02.2=0,04\left(g\right)\)
\(V_{H_2}=n.22,4=0,02.22,4=0,448\left(l\right)\)
b. \(m_{CO_2}=0,25.44=11\left(g\right)\)
\(V_{CO_2}=0,25.22,4=5,6\left(l\right)\)
c. \(m_{CO_2}=0,01.44=0,44\left(g\right)\)
\(V_{CO_2}=0,01.22,4=0,224\left(l\right)\)
d. \(m_{hhKhi}=\left(0,4.32\right)+\left(0,15.46\right)=19,7\left(g\right)\)
\(V_{hhKhi}=\left(0,4.22,4\right)+\left(0,15.22,4\right)=12,32\left(l\right)\)
2Al + 6HCl → 2AlCl3 + 3H2
a) Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}\times0,2=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,3\times22,4=6,72\left(l\right)\)
b) Theo PT: \(n_{Al}pư=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}\times0,15=0,1\left(mol\right)\)
\(\Rightarrow H=\dfrac{n_{Al}pư}{n_{Al}}\times100\%=\dfrac{0,1}{0,2}\times100\%=50\%\)
PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
\(2Zn+O_2\underrightarrow{t^o}2ZnO\)
Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}+\dfrac{1}{2}n_{Zn}=\dfrac{3}{4}x+\dfrac{1}{2}y\left(mol\right)\)
\(\Rightarrow\dfrac{3}{4}x+\dfrac{1}{2}y=0,15\)