1.a lot 

2. a little

\(n_{Mg}=\dfrac{2.4}{24}=0.1\left(mol\right)\)

\(2Mg+O_2\underrightarrow{^{^{t^0}}}2MgO\)

\(0.1........0.05........0.1\)

\(m_{MgO}=0.1\cdot40=4\left(g\right)\)

\(V_{O_3}=0.05\cdot22.4=1.12\left(l\right)\)

\(n_{KOH}=0.15\cdot2=0.3\left(mol\right)\)

\(n_{K_2CO_3}=\dfrac{1.38}{138}=0.01\left(mol\right)\)

TH1 : Chỉ sinh ra K₂CO₃, KOH dư 

\(2KOH+CO_2\rightarrow K_2CO_3+H_2O\)

\(0.01..........0.01............0.01\)

\(V=0.01\cdot22.4=0.224\left(l\right)\)

TH2 : Sinh ra 2 muối K₂CO₃, KHCO₃

\(2KOH+CO_2\rightarrow K_2CO_3+H_2O\)

\(KOH+CO_2\rightarrow KHCO_3\)

\(n_{KHCO_3}=n_{KOH}-2n_{K_2CO_3}=0.3-2\cdot0.01=0.28\left(mol\right)\)

\(n_{CO_2}=0.01+0.28=0.29\left(mol\right)\)

\(V=0.29\cdot22.4=6.496\left(l\right)\)

\(\Rightarrow0.224\left(l\right)\le V_{CO_2}< 6.496\left(l\right)\)

C. He wishes he didn't have to go now 

C. I wish he helped his mother with house work

\(n_{H_2}=\dfrac{33.6}{22.4}=1.5\left(mol\right)\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(1.5........1.5.....................1.5\)

\(m_{Fe}=1.5\cdot56=84\left(g\right)\)

\(C_{M_{H_2SO_4}}=\dfrac{1.5}{0.5}=3\left(M\right)\)

\(2H_2+O_2\underrightarrow{^{t^0}}2H_2O\)

\(1.5........0.75\)

\(V_{O_2}=0.75\cdot22.4=16.8\left(l\right)\)

\(n_{Al}=\dfrac{8.1}{27}=0.3\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(0.3..........0.9...........0.3..........0.45\)

\(C_{M_{HCl}}=\dfrac{0.9}{0.1}=9\left(M\right)\)

\(m_{AlCl_3}=0.3\cdot133.5=40.05\left(g\right)\)

\(V_{H_2}=0.45\cdot22.4=10.08\left(l\right)\)

\(n_{Al}=\dfrac{8.1}{27}=0.3\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(0.3..........0.9...........0.3..........0.45\)

\(C_{M_{HCl}}=\dfrac{0.9}{0.1}=9\left(M\right)\)

\(m_{AlCl_3}=0.3\cdot133.5=40.05\left(g\right)\)

\(V_{H_2}=0.45\cdot22.4=10.08\left(l\right)\)

\(n_{Zn}=\dfrac{19.5}{65}=0.3\left(mol\right)\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(0.3........................0.3.......0.3\)

\(m_{ZnSO_4}=0.3\cdot161=48.3\left(g\right)\)

\(V_{H_2}=0.3\cdot22.4=6.72\left(l\right)\)

\(^{C_M}Ba\left(NO_3\right)_2=0.02\left(M\right)\)

\(\left[Ba^{2+}\right]=0.02\left(M\right)\)

\(\left[NO_3^-\right]=0.02\cdot2=0.04\left(M\right)\)