\(a.m_{CO_2}=3.44=132\left(g\right)\\ m_{CO}=2.28=56\left(g\right)\\b. m_{SO_2}=0,1.64=64\left(g\right)\\ m_{O_2}=0,05.32=16\left(g\right)\)

thx u very much...friend :3

\(\overline{M}=\dfrac{0.5\cdot44+1.5\cdot32}{0.5+1.5}=35\left(\dfrac{g}{mol}\right)\)

1. Ta có: n_SO2 = \(\frac{2,24}{22,4}=0,1\left(mol\right)\)

=> m_SO2 = 0,1 x 64 = 6,4 (gam)

n_O2 = \(\frac{3,36}{22,4}=0,15\left(mol\right)\)

=> m_O2 = 0,15 x 32 = 4,8 (gam)

=> Khối lượng hỗn hợp:m_{hỗn hợp}=m_O2 + m_SO2 = 6,4 + 4,8 = 11,2g

2. Ta có: n_CO2 = \(\frac{4,4}{44}=0,1\left(mol\right)\)

n_O2 = \(\frac{3,2}{32}=0,1\left(mol\right)\)

=> V_{hỗn hợp}(đktc) = ( 0,1 + 0,1 ) x 22,4 = 4,48 (l)

3. Số mol H₂O: n_H2O = \(\frac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\)

Câu 3:

\(n_{H_2O}=\frac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\)

a, V_{O\(_2\)} = 0,15 . 22,4 = 3,36 lít

b, V_{\(CO_2\)}  = \((\dfrac{48}{44}).22,4\approx24,43\) ( lít )

c, \(V_{SO_2}=\left(\dfrac{16}{64}\right).22,4=5,6\) ( lít )

\(V_{H_2}=\left(\dfrac{18.10^{23}}{6.10^{23}}\right).22,4=67,2\) ( lít )

=> \(V_{hh}=5,6+67,2=72,8\) ( lít )

Bài 31: 

a, m_{Fe\(_2\)O\(_3\)} = 0,4. 160 = 64( g )

b, \(m_{CO_2}=\left(\dfrac{14,56}{22,4}\right).44=28,6\) ( g )

c, \(m_{O_2}=\left(\dfrac{1,2.10^{23}}{6.10^{23}}\right).32=6,4\) ( g )

\(a.n_{CO_2}=\dfrac{m_{CO_2}}{M_{CO_2}}=\dfrac{4,4}{44}=0,1\left(mol\right)\\ \Rightarrow V_{CO_2}=n_{CO_2}.22,4=0,1.22,4=2,24\left(l\right)\\ n_{O_2}=\dfrac{m_{O_2}}{M_{O_2}}=\dfrac{3,2}{32}=0,1\left(mol\right)\) 

\(\Rightarrow V_{O_2}=n_{O_2}.22,4=0,1.22,4=2,24\left(l\right)\) 

\(n_{H_2}\)  = 2mol

=>\(V_{H_2}\) = 2. 22,4 = 44,8l

\(n_{O_2}\) = 0,0875 mol

=>\(V_{O_2}\) = 0,0875 . 22,4 = 1,96l

\(n_{CO_2}\) = 0,5 mol

=>\(V_{CO_2}\)  = 0,5 .22,4 = \(11,2\left(l\right)\) 

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

=>\(V_{O_2}\) = 0,2 . 22,4 = \(4,48\left(l\right)\) 

b) Ta có:

\(n_{NH_3}=\dfrac{6.10^{23}}{6.10^{23}}=1\left(mol\right)\\ n_{O_2}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\\ \Sigma n_{hh}=1+0,5=1,5\left(mol\right)\\ V_{hh}=1,5.22,4=33,6\left(l\right)\)

a, Ta có : n_SO2 = 2,24 / 22,4 = 0,1 (mol)

=> m_SO2 = n * M = 0,1 * ( 32+16*2)= 6,4 (g)

n_O2 = 3,36 / 22,4 = 0,15 (mol)

=> m_O2 = 0,15 * 32 = 4,8 (g)

b, Ta có: n_CO2 = m / M = 4,4 / (12 +16*2) = 0,1 (mol)

=> V_CO2 = n* 22,4 =0,1 *22,4= 2,24 (lít)

n_O2 = 3,2 / 32= 0,1 (mol)

=> V_O2 = 0,1 *22,4 =2,24 (lít)

c, n_N2 = 4,48 / 22,4 = 0,2 (mol)

d, m_H2O = 0,5 * (1*2+16) = 9 (g)

n_H2SO4 = 19,6 / (1*2+32+16*4)= 3,2 (mol)

e, M _{h/c A} = 8 /0,2=40(g)

f, 1. n_O2 = 8/32= 0,25 (mol)

2. số phân tử oxi là : 0,25 *6 *10²³ = 1,5*10²³ (p. tử)

3. V_O2 = 0,25 * 22,4 =5,6 (lít)

Câu c bạn làm cụ thể hơn được không ???

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