a: \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)
b: \(n_{CO_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)
\(\Leftrightarrow n_{H_2O}=2\cdot0.15=0.3\left(mol\right)\)
\(\Leftrightarrow n_{NaOH}=0.15\left(mol\right)\)
\(m_{NaOH}=0.15\cdot40=6\left(g\right)\)
a) 2NaOH + CO2 --> Na2CO3 + H2O
b) \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: 2NaOH + CO2 --> Na2CO3 + H2O
______0,3<---0,15------->0,15------>0,15
=> mNaOH = 0,3.40 = 12 (g)
c) msp = 0,15.106 + 0,15.18 = 18,6(g)