Đáp án:
Giải thích các bước giải:
$n_{NaOH} = 0,2.0,75 =0,15(mol)$
$n_{CO_2} = \dfrac{2,24}{22,4} = 0,1(mol)$
Ta thấy :
$1 < \dfrac{n_{NaOH}}{n_{CO_2}} = \dfrac{0,15}{0,1} = 1,5 < 2$
Do đó, muối sau phản ứng gồm : $NaHCO_3(a\ mol) ; Na_2CO_3(b\ mol)$
$2NaOH + CO_2 \to Na_2CO_3 + H_2O$$NaOH + CO_2 \to NaHCO_3$Theo PTHH :
$n_{CO_2} = a + b = 0,1(mol)$$n_{NaOH} = a + 2b = 0,15(mol)$$\to = a = 0,05 ; b = 0,05$Vậy :$m_{NaHCO_3} = 0,05.84 = 4,2(gam)$$m_{Na_2CO_3} = 0,05.106 = 5,3(gam)$
-dung dịch xút là :dd NaOH
n CO2=$\frac{2,24}{22,4}$=0,1 mol
n NaOH=0,2.0,75=0,15 mol
-Ta có:1 <T=$\frac{nNaOH}{nCO2}$=$\frac{0,15}{0,1}$=1,5 <2
⇒Tạo 2 muối
-gọi n NaHCO3=x mol
n Na2CO3=y mol
CO2+NaOH→NaHCO3
x x ←x mol
CO2+2NaOH→Na2CO3+H2O
y 2y ←y mol
-Ta có hệ pt: $\left \{ {{x+y=0,1} \atop {x+2y=0,15}} \right.$ $\left \{ {{x=0,05} \atop {y=0,05}} \right.$
⇒m muối=m NaHCO3+m Na2CO3
=0,05.84+0,05.106=9,5 g
------------------Ngyễn Hoạt------------------
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