Đáp án:
M là Canxi
$\eqalign{
& \% {m_{CaC{O_3}}} = 16\% \cr
& \% {m_{CaO}} = 84\% \cr} $
Giải thích các bước giải:
MO + H2SO4 → MSO4 + H2O (1)
MCO3 + H2SO4 → MSO4 + CO2↑ + H2O (2)
B: CO2
${n_{C{O_2}}} = {{7,04\% m} \over {44}} = 0,16\% m$
${n_{MS{O_4}}} = {{225,76\% m} \over {{M_M} + 96}} = {{2,2576m} \over {{M_M} + 96}}mol$
Theo (2): nMCO3 = nMSO4 (2) = nCO2 = ${n_{C{O_2}}} = {{7,04\% m} \over {44}} = 0,0016m$
mA = mMO + mMCO3
$\eqalign{
& \to {m_{MO}} = m - 0,0016m.({M_M} + 60) = 0,904m - 0,0016m.{M_M} \cr
& {n_{MO}} = {{0,904m - 0,0016m.{M_M}} \over {{M_M} + 16}} \cr} $
Theo (1) : ${n_{MS{O_4}(1)}} = {n_{MO}} = {{0,904m - 0,0016m.{M_M}} \over {{M_M} + 16}}$
$\eqalign{
& {n_{MS{O_4}}} = {n_{MS{O_4}(1)}} + {n_{MS{O_4}(2)}} \cr
& \to {{2,2576m} \over {{M_M} + 96}} = {{0,904m - 0,0016m.{M_M}} \over {{M_M} + 16}} + 0,0016m \cr
& \to {M_M} = 40 \cr} $
Oxit: CaO
Muối cacbonat: CaCO3
nCaCO3 = 0,0016m (mol)
→ mCaCO3 = 0,0016m . 100 = 0,16m (g)
$\eqalign{
& \% {m_{CaC{O_3}}} = {{0,16m} \over m}.100\% = 16\% \cr
& \% {m_{CaO}} = 100\% - 16\% = 84\% \cr} $
