Đáp án:

$\rm V=3,36(l)$

$\rm C$`%`$\rm _{dd\ CaCl_2}$`~~15,36%\ , ` $\rm C$`%`$\rm _{dd\ HCl_{(dư)}}$`~~3,37%`

Giải thích các bước giải:

            $\rm CaCO_3+2HCl\to CaCl_2+CO_2+H_2O$

$\rm a)$

$\rm n_{CaCO_3}=\dfrac{15}{100}=0,15(mol)$

$\rm n_{HCl}$`={100.14,6%}/{36,5}=`$\rm 0,4(mol)$

$\rm Ta\ có : n_{CaCO_3}<\dfrac{n_{HCl}}{2}$`=>`$\rm HCl\ dư$

$\rm n_{CO_2}=n_{CaCO_3}=0,15(mol)$

$\rm V_{CO_2}=0,15.22,4=3,36(l)$

$\rm b)$

$\rm n_{HCl(pu)}=2n_{CaCO_3}=0,3(mol)$

$\rm n_{CaCl_2}=n_{CaCO_3}=0,15(mol)$

$\rm m_{dd\ spu}=15+100-(0,15.44)=108,4(g)$

$\rm C$`%`$\rm _{dd\ CaCl_2}$`={0,15.111}/{108,4}.100%~~15,36%`

$\rm C$`%`$\rm _{dd\ HCl_{(dư)}}$`={(0,4-0,3).36,5}/{108,4}.100%~~3,37%`

PTHH: CaCOPTHH: $CaCO_{3}$ + $2HCl_{}$ -> $CO_{2}$ + $CaCl_{2}$ +$H_{2}O$

$n_{CaCO}$ $_{3}$ = $\frac{15}{100}$ $0,15^{}$ $(mol)_{}$

$n_{HCl}$ = $\frac{100 . 16,4%}{36,5}$ = 0,4 mol

Ta có: $n_{CaCO}$ $_{3}$ < $\frac{nHCl}{2}$ => HCl dư.

a) nCO$_{2}$ = nCaCO$_{3}$ = 0,15 (mol)

VCO2 = 0,15 . 22,4 = 3,36 (l)

b) nHCl = 2nCaCO3 = 0,3 (mol)

nCaCl2 = nCaCO3 = 0,15

$m_{dd}$= 15 + 100 - ( 0,15 . 44) = 108,4 (gam)

C%$_{ddCaCl2}$ $_{}$ =$\frac{(0,15 . 111}{108,4}$ ≈15,3%

C%$_{ddHCl (dư)}$ $_{}$ =$\frac{(0,4 - 0,3 . 36,5}{108,4}$ ≈3,37%

~chúc bạn học tốt cho mình 5* + clthn nhé~