Question

Answer 1

C8 . 

b. \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

\(SO_2+2NaOH\rightarrow Na_2SO_3+H_2O\)

\(nhhA=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

=>\(nNaOH=0,1.2=0,2\left(mol\right)\)

Từ ct : \(CM=\dfrac{n}{V}\)

=> \(VNaOH=\dfrac{nNaOH}{CM_{NaOH}}=\dfrac{0,2}{1}=0,2\left(l\right)\)

Answer 2

C8:

\(nA=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

\(MA=24.2=48g/mol\)

\(\Rightarrow mA=0,5.48=24\left(g\right)\)

=>\(mCO_2+mSO_2=24\left(g\right)\)

<=>\(44.nCO_2+64.nSO_2=24\left(g\right)\)

=> nCO2 = 0,2225 (mol)

=> nSO2 = 0,22203125 (mol)

Answer 3

c9 bn cắt quá k thấy khúc dưới lm lun bn coi lại hén