\(n_{H_2}=\dfrac{12.10^{22}}{6.10^{23}}=0,2\left(mol\right)\)

\(n_{SO_2}=\dfrac{3,2}{64}=0,05\left(mol\right)\)

=> V_hh = 0,25.22,4 + 4,48 + 0,2.22,4 + 0,05.22,4 = 15,68(l)

\(V_{O_2}=0,25.22,4=5,6l\)

\(V_{H_2}=\left(\dfrac{1,2.10^{23}}{6.10^{23}}\right).22,4=4,48l\)

\(V_{SO_2}=\left(\dfrac{3,2}{64}\right).22,4=1,12l\)

=> V_hh = 5,6 + 4,48 + 4,48 + 1,12 = 15,68 lít

n_{O (H2O)} = n_H2O = 0,25 mol

n_{O (CO2)} = 2n_CO2 = 2 . 1,5 = 3 mol

n_{O (SO2)} = 2n_SO2 = 2 . 0,75 = 1,5 mol

\(\sum n_O=0,25+3+1,5=4,75\left(mol\right)\)

Số nguyên tử O = \(4,75\times6\times10^{23}=28,5\times10^{23}\)

\(m_{O2}=n_{O2}.M_{O2}=0,1.32=3,2\left(g\right)\)

\(m_{H2}=n_{H2}.M_{H2}=0,15.2=0,3\left(g\right)\)

\(m_{SO2}=n_{SO2}.M_{SO2}=0,25.64=16\left(g\right)\)

\(m_{hh}=m_{O2}+m_{H2}+m_{SO2}=19,2\left(g\right)\)

P/s: Ko biết lm thế này có đc khum nhỉ ??

Bài 1:

a) \(V_{khí}=\left(0,2+0,5+0,35\right)\cdot22,4=23,52\left(l\right)\)

b) \(m_{khí}=0,2\cdot64+0,5\cdot28+0,35\cdot28=36,6\left(g\right)\)

Bài 2:

Sửa đề: Tỉ lệ \(Mg:C:O=2:1:4\)

Ta có: \(n_{Mg}:n_C:n_O=\dfrac{2}{24}:\dfrac{1}{12}:\dfrac{4}{16}=1:1:3\)

\(\Rightarrow\) CTHH là MgCO₃

\(a.n_{H_2}=\dfrac{1,2.10^{23}}{6.10^{23}}=0,2\left(mol\right)\\ n_{SO_2}=\dfrac{6,4}{64}=0,1\left(mol\right)\\ V_X=\left(1,5+2,5+0,2+0,1\right).22,4=96,32\left(l\right)\\b. m_X=1,5.32+2,5.28+0,2.2+6,4=124,8\left(g\right)\)

a.nH2=1,2.10236.1023=0,2(mol)nSO2=6,464=0,1(mol)VX=(1,5+2,5+0,2+0,1).22,4=96,32(l)b.mX=1,5.32+2,5.28+0,2.2+6,4=124,8(g)

Bài 2:

a) Vì khối lượng mol của N₂ và CO đều bằng 28 và lớn hơn khối lượng mol của khí metan CH₄ (28>16)

=> \(d_{\dfrac{hhX}{CH_4}}=\dfrac{28}{16}=1,75\)

Hỗn hợp X nhẹ hơn không khí (28<29)

b)

 \(M_{C_2H_4}=M_{N_2}=M_{CO}=28\left(\dfrac{g}{mol}\right)\\ \rightarrow M_{hhY}=28\left(\dfrac{g}{mol}\right)\\ d_{\dfrac{Y}{H_2}}=\dfrac{28}{2}=14\)

c) \(\%V_{NO}=100\%-\left(30\%+30\%\right)=40\%\\ \rightarrow\%n_{CH_4}=40\%\\ Vì:\%m_{CH_4}=22,377\%\\ Nên:\dfrac{30\%.16}{40\%.30+30\%.16+30\%.\left(x.14+16\right)}=22,377\%\\ \Leftrightarrow x=-0,03\)

Sao lại âm ta, để xíu anh xem lại như nào nhé. 

Bài 1:

\(a.\\ d_{\dfrac{SO_2}{O_2}}=\dfrac{64}{32}=2\\ d_{\dfrac{SO_2}{N_2}}=\dfrac{64}{28}=\dfrac{16}{7}\\ d_{\dfrac{SO_2}{SO_3}}=\dfrac{64}{80}=0,8\\ d_{\dfrac{SO_2}{CO}}=\dfrac{64}{28}=\dfrac{16}{7}\\ d_{\dfrac{SO_2}{N_2O}}=\dfrac{64}{44}=\dfrac{16}{11}\\ d_{\dfrac{SO_2}{NO_2}}=\dfrac{64}{46}=\dfrac{32}{23}\\ b.M_{hhA}=\dfrac{1.64+1.32}{1+1}=48\left(\dfrac{g}{mol}\right)\\ d_{\dfrac{hhA}{O_2}}=\dfrac{48}{32}=1,5\)

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

\(a_1,m_{CaCO_3}=0,25.100=25(g)\\ a_2,m_{SO_2}=\dfrac{3,36}{22,4}.64=9,6(g)\\ a_3,m_{H_2SO_4}=\dfrac{9.10^{23}}{6.10^{23}}.98=147(g)\)