Mg + 2HCl \(\rightarrow\)MgCl2 + H2
nMg=\(\dfrac{2,4}{24}=0,1\left(mol\right)\)
Theo pTHH ta có;
nH2=nMg=0,1(mol)
VH2=22,4.0,1=2,24(lít)
2Na + Cl2 \(\rightarrow\)2NaCl
nCl2=\(\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PTHH ta có:
2nCl2=nNa=0,2(mol)
mNa=23.0,2=4,6(g)
2NaOH + H2SO4 \(\rightarrow\)Na2SO4 + H2O
nH2SO4=0,1.1=0,1(mol)
Theo PTHH ta có:
nNaOH=2nH2SO4=0,2(mol)
Vdd NaOH=\(\dfrac{0,2}{1}=0,2\left(lít\right)\)