a)

Ban đầu (n_{M_2CO_3};n_{MHCO_3};n_{MCl})=(x;y;z)(mol)

Bảo toàn C: n_{CO_2}=x+y={17,6}/{44}=0,4(1)

Bảo toàn M: n_{MCl(B)}=2x+y+z(mol)

->n_{MCl(1P)}=x+0,5y+0,5z(mol)

B gồm HCl dư, MCl

P1:

HCl+KOH->KCl+H_2O

->n_{HCl(1P)}=n_{KOH}=0,125.0,8=0,1(mol)

P2:

HCl+AgNO_3->AgCl+HNO_3

MCl+AgNO_3->MNO_3+AgCl

->n_{AgCl}=0,1+x+0,5y+0,5z={68,88}/{143,5}=0,48

->2x+y+z=0,76(2)

(1)(2)->2x+2y-2x-y-z=0,8-0,76=0,04

->z=y-0,04

(1)->x=0,4-y

m_X=x(2M+60)+y(M+61)+z(M+35,5)=43,71

->M(2x+y+z)+60x+61y+35,5z=43,71

->0,76M+60(0,4-y)+61y+35,5(y-0,04)=43,71

->0,76M+22,58+36,5y=43,71

->M=27,8-48,03y

y>0;y<0,25

->15,8

->M=23

->M:\ Na

b)

Có hệ \(\begin{cases} 106x+84y+58,5z=43,71\\ x+y=0,4\\ 2x+y+z=0,76\end{cases}\)

->x=0,3;y=0,1;z=0,06

->\%m_{Na_2CO_3}=72,75\%;\%m_{NaHCO_3}=19,22\%;\%m_{NaCl}=8,03\%

c)

n_{HCl\ du}=0,1.2=0,2(mol)

n_{HCl\ pu}=2x+y=0,7(mol)

->m_{dd\ HCl}={(0,7+0,2).36,5}/{10,52\%}={82125}/{263}(g)

->V={{82125}/{263}}/{1,05}\approx 297,39(mL)

n_{KCl}=n_{KOH}=0,1(mol);n_{NaCl(1P)}=0,3+(0,1+0,06).0,5=0,38(mol)

->m=0,1.74,5+0,38.58,5=29,68(g)