a) mO2= nO2. M(O2)=0,45. 32=14,4(g)

b) mBaCO3=nBaCO3.M(BaCO3)=0,6.197=118,2(g)

c) mAl2(SO4)3=nAl2(SO4)3.M(Al2(SO4)3)=1,5.342=513(g)

d) nSO2=V(SO2,đktc)/22,4=16,8/22,4=0,7(mol)

=> mSO2=nSO2.M(SO2)=0,7.64=44,8(g)

e) nH2O=(3.10²³):(6.10²³)=0,5(mol)

=>mH2O=nH2O.M(H2O)=0,5.18=9(g)

f) nCO2=V(CO2,đktc)/22,4=8,96/22,4=0,4(mol)

=>mCO2=nCO2.M(CO2)=0,4.44=17,6(g)

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 )

4.

a) \(V_{SO_2}=0.5\cdot22.4=11.2\left(l\right)\)

b) \(V_{CH_4}=\dfrac{3.2}{16}\cdot22.4=4.48\left(l\right)\)

c) \(V_{N_2}=\dfrac{0.9\cdot10^{23}}{6\cdot10^{23}}\cdot22.4=3.36\left(l\right)\)

5.

a) \(m_{Al}=0.1\cdot27=2.7\left(g\right)\)

b) \(m_{Cu\left(NO_3\right)_2}=0.3\cdot188=56.4\left(g\right)\)

c) \(m_{Na_2CO_3}=\dfrac{1.2\cdot10^{23}}{6\cdot10^{23}}\cdot106=21.2\left(g\right)\)

d) \(m_{CO_2}=\dfrac{8.96}{22.4}\cdot44=17.6\left(g\right)\)

e) \(m_K=0.5\cdot2\cdot39=39\left(g\right)\\ m_C=0.5\cdot12=6\left(g\right)\\ m_O=0.5\cdot3\cdot16=24\left(g\right)\)

\(a.\)

\(m_{CuO}=2\cdot40=80\left(g\right)\)

\(b.\)

\(n_{CO_2}=\dfrac{8.96}{22,4}=0.4\left(mol\right)\)

\(c.\)

\(n_{CH_4}=\dfrac{1.2\cdot10^{23}}{6\cdot10^{23}}=0.2\left(mol\right)\)

\(V_{CH_4}=0.2\cdot22.4=4.48\left(l\right)\)