1)
\(n_{CO2}=0,2\left(mol\right)\)
\(n_{NaOH}=2\left(mol\right)=n_{OH^-}\)
\(\frac{n_{OH^-}}{n_{CO2}}=15\rightarrow\) Chỉ tạo muối cacbonat, kiềm dư
\(PTHH:2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)
\(\rightarrow n_{NaOH_{pu}}=0,4\left(mol\right),n_{Na2CO3}=0,2\left(mol\right)\)
\(n_{NaOH_{du}}=3-0,4=2,6\left(mol\right)\)
\(\rightarrow m=2,6.40+0,2.106=125,2\left(g\right)\)
2) \(n_{CO2}=0,3\left(mol\right),n_{OH^-}=0,5\left(mol\right)\)
\(\frac{n_{OH^-}}{n_{CO2}}=1,6\rightarrow\) Tạo 2 muối
\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)
\(Ca\left(OH\right)_2+2CO_2\rightarrow Ca\left(HCO_3\right)_2\)
Gọi a là mol CaCO3, b là mol Ca(HCO3)2 \(\rightarrow\left\{{}\begin{matrix}a+b=0,25\\a+2b=0,3\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,05\end{matrix}\right.\)
\(\rightarrow m_{CaCO3}=100.0,2=20\left(g\right)\)
Câu 1 :
nCO2 = 0.2 mol
nNaOH = 0.24 mol
nNaOH/nCO2 = 0.24/0.2 = 1.2 => tạo ra 2 muối
Đặt : nNa2CO3 = x mol
nNaHCO3 =y mol
<=> 2x + y = 0.24
x + y = 0.2
=> x = 0.04
y= 0.16
mM = m = 0.04*106+0.16*84=17.68 g
Câu 2 :
nCO2 = 0.3 mol
nCO2/nCa(OH)2 = 0.3/0.25= 1.2 => Tạo ra 2 muối
Đặt :
nCaCO3 = x mol
nCa(HCO3)2 = y mol
<=> x + y = 0.25 mol
x + 2y = 0.3
=> x = 0.2
y = 0.05
mCaCO3 = 0.2*100=20 g
