nNaOH = 6/40 = 0,15

=> CM NaOH = 0,15/ 1,5 = 0,1M

Ta có: \(C\%_{NaOH}=\dfrac{8}{8+100}.100\%\approx7,41\%\)

\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)

V _{dd sau hòa tan} = 100 (ml) = 0,1 (l)

\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,2}{0,1}=2\left(M\right)\)

Mình đã giúp bạn ở dưới rồi nhé :)

\(n_{CaO}=\dfrac{3.92}{56}=0.07\left(mol\right)\)

\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

\(0.07.....................0.07\)

\(n_{CO_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)

\(T=\dfrac{0.05}{0.07}=0.7\)

\(\Rightarrow\text{Tạo ra muối TH}\)

\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)

\(0.05...............0.05.........0.05\)

\(m_{CaCO_3}=0.05\cdot100=5\left(g\right)\)

\(C_{M_{Ca\left(OH\right)_2\left(dư\right)}}=\dfrac{0.07-0.05}{0.7}=\dfrac{1}{35}\left(M\right)\)

\(CuO+2HCl\rightarrow CuCl_2+H_2O\\ n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\\ n_{HCl}=2.0,1=0,2\left(mol\right)\\ V_{ddHCl}=\dfrac{0,2}{1}=0,2\left(l\right)\\ V_{ddsau}=V_{ddHCl}=0,2\left(l\right)\\ n_{CuCl_2}=n_{CuO}=0,2\left(mol\right)\\ C_{MddCuCl_2}=\dfrac{0,2}{0,2}=1\left(M\right)\)

Chọn đáp án C

PTHH: \(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

Ta có: \(\left\{{}\begin{matrix}n_{Ca\left(OH\right)_2}=n_{CaO}=\dfrac{3,92}{56}=0,07\left(mol\right)\\n_{CO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Tạo muối trung hòa

PTHH: \(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O\)

Ta có: \(n_{Ca\left(OH\right)_2\left(dư\right)}=0,07-n_{Ca\left(OH\right)_2\left(p/ứ\right)}=0,07-n_{CO_2}=0,02\left(mol\right)\)

\(\Rightarrow C_{M_{Ca\left(OH\right)_2\left(dư\right)}}=\dfrac{0,02}{0,7}\approx0,03\left(M\right)\)