\(n_{CO_2}=\dfrac{25,2}{22,4}=1,125mol\)
\(n_{NaOH}=0,9.2=1,8mol\)
\(1< \dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{1,8}{1,125}=1,6< 2\)\(\rightarrow\)Tạo 2 muối Na2CO3(xmol) và NaHCO3(ymol)
CO2+2NaOH\(\rightarrow\)Na2CO3+H2O
CO2+NaOH\(\rightarrow\)NaHCO3
Ta có hệ:
x+y=1,125
2x+y=1,8
Giải ra x=0,675 và y=0,45
Tổng khối lượng muối trong A:
m=0,675.106+0,45.84=109,35g
G=\(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\frac{x^2-2x+1}{2}\)
\(=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\frac{\left(\sqrt{x-1}\right)^2}{2}\)
\(=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)^{ }-\left(\sqrt{x+2}\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right)\)\(.\frac{\left(\sqrt{x}-1\right)^2}{2}\)
\(=\left(\frac{x+\sqrt{x}-2\sqrt{x}-2-x+\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\frac{\sqrt{x}-1}{2}\)
\(=\frac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)^2}.\frac{\sqrt{x}-1}{2}=\frac{-x-\sqrt{x}}{\left(\sqrt{x}+1\right)^2}\)
\(=\frac{-\left(x+\sqrt{x}\right)}{\left(\sqrt{x}+1\right)^2}=\frac{-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2}=\frac{-\sqrt{x}}{\sqrt{x}+1}\)
