Đáp án:
Vậy: $12,5≤m≤30$
Giải thích các bước giải:
$3,36≤V≤10,64⇒0,15≤n_{CO_2}≤0,475$
$n_{Ca(OH)_2}=1.0,3=0,3\ mol$
$⇒4≥\dfrac{2.n_{Ca(OH)_2}}{n_{CO_2}}≥1,263$
TH1:
$2>\dfrac{2.n_{Ca(OH)_2}}{n_{CO_2}}≥1,263⇔1>\dfrac{n_{Ca(OH)_2}}{n_{CO_2}}≥0,631$
Hay: $0,3≤<n_{CO_2}<0,475$
PTHH:
$CO_2+Ca(OH)_2\to CaCO_3+H_2O$
$CO_2+ CaCO_3+H_2O\to Ca(HCO_3)_2$
$n_↓=2n_{Ca(OH)_2}-n_{CO_2}⇔2.0,3-0,475<n_↓≤0,6-0,3\\⇔0,125≤n_↓<0,3$
$⇔12,5≤m<30$
TH2: $2≤\dfrac{2.n_{Ca(OH)_2}}{n_{CO_2}}≤4$
Hay: $0,15≤n_{CO_2}≤0,3$
$CO_2+Ca(OH)_2\to CaCO_3+H_2O$
$n_↓=n_{CO_2}⇒0,15≤n_↓≤0,3⇒15≤m≤30$
Vậy: $12,5≤m≤30$
