Question

Answer 1

n(peptit)=0,06(mol)

PTHH: Gly-Ala-Ala + 3 KOH -> Gly-K + 2 Ala-K + H2O

nKOH=0,2(mol)

=> m(rắn)= m(muối)+ mKOH(dư)= (13,02+ 0,18.56- 0,06.18)+0,02.56=23,14(g)

=>CHỌN B

Answer 2

\(n_A=\dfrac{13.02}{217}=0.06\left(mol\right)\)

\(\Rightarrow n_{H_2O}=\dfrac{0.2}{3}=\dfrac{1}{15}\left(mol\right)\)

\(n_{KOH}=0.2\cdot1=0.2\left(mol\right)\)

\(\Rightarrow n_{KOH\left(dư\right)}=0.2-0.06\cdot3=0.02\left(mol\right)\)

\(BTKL:\)

\(m_{peptit}+m_{KOH}=m_r+m_{H_2O}\)

\(\Rightarrow m_r=13.02+0.2\cdot56-0.06\cdot18+0.02\cdot56=23.14\left(g\right)\)

\(\)