\(n_{CO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(m_{ddNaOH}=1,3.500=650\left(g\right)\)
\(\dfrac{m_{ctNaOH}.100\%}{650}=25\%\Rightarrow m_{ctNaOH}=162,5\left(g\right)\)
\(n_{NaOH}=\dfrac{162,5}{40}=4,0625\left(mol\right)\)=>NaOH dư
\(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)
0,5-----------------------0,5---------------