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, 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 )

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

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

=> 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)

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

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

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

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

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

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

=> \(m_{hh}=48+70+0,4+6,4==124,8g\)

a.

\(m_{Al}=0.5\cdot27=13.5\left(g\right)\)

\(m_{CO_2}=\dfrac{6.72}{22.4}\cdot44=13.2\left(g\right)\)

\(m_{N_2}=\dfrac{5.6}{22.4}\cdot28=7\left(g\right)\)

\(m_{CaCO_3}=0.25\cdot100=25\left(g\right)\)

b.

\(m_{hh}=\dfrac{3.36}{22.4}\cdot2+\dfrac{5.6}{22.4}\cdot28+0.2\cdot44=16.1\left(g\right)\)

e cảm ơn^^

a) nCO2=[(9.10²³)/(6.10²³)]=1,5(mol)

=> mCO2=1,5.44=66(g)

V(CO2,đktc)=1,5.22,4=33,6(l)

b) nH2=4/2=2(mol)

N(H2)=2.6.10²³=12.10²³(phân tử)

V(H2,đktc)=2.22,4=44,8(l)

c) N(CO2)=0,5.6.10²³=3.10²³(phân tử)

V(CO2,đktc)=0,5.22,4=11,2(l)

mCO2=0,5.44=22(g)

d) nN2=2,24/22,4=0,1(mol)

mN2=0,1.28=2,8(g)

N(N2)=0,1.10²³.6=6.10²² (phân tử)

e) nCu=[(3,01.10²³)/(6,02.10²³)]=0,5(mol)

mCu=0,5.64=32(g)

Mà sao tính thể tích ta :3

mMg = 0,5.24 = 12 gam

V_SO2 = n.22,4 = 0,25.22,4 = 5,6 lít

nN₂ = \(\dfrac{16,8}{22,4}\)= 0,75 mol , nO₂ = \(\dfrac{5,6}{22,4}\)= 0,25 mol

=> m(N₂  + O₂ ) = 0,75.28 + 0,25.32 = 29 gam