\(a)m_{NaOH}=\dfrac{m_{d^2}\cdot C\%}{100}=\dfrac{200\cdot10}{100}=20\left(g\right)\\ \Rightarrow n_{NaOH}=\dfrac{m}{M}=\dfrac{20}{40}=0,5\left(mol\right)\)
Để sản phẩm tạo muối trung hòa
thì \(\Rightarrow\dfrac{n_{NaOH}}{n_{CO_2}}\ge2\)
\(\Rightarrow\dfrac{0,5}{n_{CO_2}}\ge\dfrac{0,5}{0,25}\\ \Rightarrow n_{CO_2}\le0,25\\ \Rightarrow m_{CO_2}=n\cdot M\le0,25\cdot44=11\left(g\right)\)
\(pthh:2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)
b) Để sản phẩm tạo muối axit
thì \(\Rightarrow\dfrac{n_{NaOH}}{n_{CO_2}}\le1\)
\(\Rightarrow\dfrac{0,5}{n_{CO_2}}\le\dfrac{0,5}{0,5}\\ \Rightarrow n_{CO_2}\ge0,5\left(mol\right)\\ \Rightarrow m_{CO_2}=n\cdot M\ge0,5\cdot44=22\left(g\right)\)
\(pthh:NaOH+CO_2\rightarrow NaHCO_3\)
c) Để sản phẩm tạo 2 muối
thì \(\Rightarrow\dfrac{n_{NaOH}}{n_{CO_2}}< 2\)
\(\Rightarrow\dfrac{0,5}{n_{CO_2}}< \dfrac{0,5}{0,25}\\ \Rightarrow n_{CO_2}>0,25\\ \Rightarrow m_{CO_2}=n\cdot M>0,25\cdot44=11\left(g\right)\)
\(\text{Mặt khác }:\dfrac{n_{NaOH}}{n_{CO_2}}>1\)
\(\Rightarrow\dfrac{0,5}{n_{CO_2}}>\dfrac{0,5}{0,5}\\ \Rightarrow n_{CO_2}< 0,5\left(mol\right)\\ \Rightarrow m_{CO_2}=n\cdot M< 0,5\cdot44=22\left(g\right)\)
\(\Rightarrow11< m_{CO_2}< 22\)
