N phân tử = 1 mol phân tử

\(\Rightarrow n_{O2}=1mol;n_{N_2}=2mol;n_{CO_2}=1,5mol\)

\(\Rightarrow m_{hh}=1.32+2.28+1,5.44=154g\)

b. \(m_{hh}=0,1.56+0,2.64+0,3.65+0,25.27=44,65g\)

c. \(n_{O_2}=\dfrac{2,24}{22,4}=0,1mol\)

\(n_{H_2}=\dfrac{1,12}{22,4}=0,05mol\)

\(n_{HCl}=\dfrac{6,72}{22,4}=0,3mol\)

\(n_{CO_2}=\dfrac{0,56}{22,4}=0,025mol\)

\(\Rightarrow m_{hh}=0,1.32+0,05.2+0,3.36,5+0,025.44=15,35g\)

khối lượng của N phân tử oxi chính là khối lượng của 1 mol oxi, ta có: 

\(m_{O_2}=16.1=16\left(g\right)\)

tương tự với \(N_2\) và \(CO_2\)ta có:

\(m_{N_2}=28.2=56\left(g\right)\)

\(m_{CO_2}=44.1,5=66\left(g\right)\)

khối lượng của hỗn hợp là:

\(m_{hh}=m_{O_2}+m_{N_2}+m_{CO_2}=16+56+66=138\left(g\right)\)

Bài 1 :

a,

- 5,6g Fe.

\(\Rightarrow n_{Fe}=\frac{5,6}{56}=0,1\left(mol\right)\)

- 4,05g Al.

\(n_{Al}=\frac{4,05}{27}=0,15\left(mol\right)\)

- 7,8g Zn

\(n_{Zn}=\frac{65}{7,8}=0,12\left(mol\right)\)

b,

b. Tính thể khí (đktc) của:

- 0,5 mol CO2.

\(V_{CO2}=0,5.22,4=11,2\left(l\right)\)

- 0,75 mol N2.

\(V_{N2}=0,75.22,4=16,8\left(l\right)\)

- 0,3 mol CO.

\(V_{CO}=0,3.22,4=6,72\left(l\right)\)

c,

- 0,5 mol CO2; 0,75 mol N2; 0,3 mol CO.

\(V_{CO2}=0,5.22,4=11,2\left(l\right),m_{CO2}=0,5.44=22\left(g\right)\)

\(V_{N2}=0,75.22,4=16,8\left(l\right);m_{N2}=0,75.14=21\left(g\right)\)

\(V_{CO}=0,3.22,4=6,72\left(l\right),m_{CO}=0,3.8,4\left(g\right)\)

- 0,25 mol CO2; 0,5 mol N2; 0,35 mol CO.

\(V_{CO2}=0,25.22,4=5,6\left(l\right);m_{CO2}=0,24.44=10,56\left(g\right)\)

\(V_{N2}=0,5.22,4=11,2\left(l\right);m_{N2}=28.0,5=9\left(g\right)\)

\(V_{CO}=0,35.22,4=7,84\left(l\right);m_{CO}=0,35.28=9,8\left(g\right)\)

- 0,05 mol CO2; 0,7 mol N2; 0,6 mol CO. (Tương tự nha )

a, \(\overline{M}=\dfrac{0,1.44+0,2.28}{0,1+0,2}\approx33,33\left(g/mol\right)\)

b, \(\overline{M}=\dfrac{0,2.28+0,3.2}{0,2+0,3}=12,4\left(g/mol\right)\)

c, \(\overline{M}=\dfrac{0,1.28+0,2.30+0,2.44}{0,1+0,2+0,2}=35,2\left(g/mol\right)\)

d, \(\overline{M}=\dfrac{0,2.56+0,1.24+0,1.27}{0,2+0,1+0,1}=40,75\left(g/mol\right)\)

Đáp án C

\(a.\)

\(m_{hh}=0.12\cdot90+0.15\cdot58=19.5\left(g\right)\)

\(b.\)

\(V_{hh}=\left(0.25+0.1+0.05\right)\cdot22.4=8.96\left(l\right)\)

\(c.\)

\(n_A=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)

\(M_A=23\cdot2=46\left(\dfrac{g}{mol}\right)\)

\(m_A=0.45\cdot46=20.7\left(g\right)\)

\(d.\)

\(n_{hh}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

Vì CO₂ : O₂ = 2 : 1 

\(\Rightarrow n_{CO_2}=0.2\left(mol\right),n_{O_2}=0.1\left(mol\right)\)

\(m_{hh}=0.2\cdot44+0.1\cdot32=12\left(g\right)\)

\(\overline{M}=\dfrac{12}{0.3}=40\left(\dfrac{g}{mol}\right)\)

Đáp án A