Question

khi cho 1,12 lít co2 -dktc- hấp thụ hết vào 100ml dung dịch naoh 0,75M thu được dung dịch chứa bn g muối

Answer 1

\(PTHH:2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)

Ban đầu :_0,075_____0,05___________________

Phứng :__0,075____0,0375________________

Sau phứng :0______0,0125____0,0375________

\(n_{CO2}=\frac{1,12}{22,4}=0,05\left(mol\right)\)

Đổi 100 ml=0,1l

\(n_{NaOH}=0,1.0,75=0,075\left(mol\right)\)

\(m_{Na2CO3}=0,0375.\left(23.2+12+16.3\right)=3,975\left(g\right)\)

Answer 2

\(n_{CO2}=\frac{1,12}{22,4}=0,05\left(mol\right)\)

\(n_{NaOH}=0,1.0,75=0,075\left(mol\right)\)

\(\frac{n_{NaOH}}{n_{CO2}}=\frac{0,075}{0,05}=1,5\)

\(\rightarrow\) Sinh ra 2 muối

\(NaOH+CO_2\rightarrow NaHCO_3\)

x _______x_____________

\(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)

y_________1/2_______________

\(\left\{{}\begin{matrix}x+y=0,075\\x+\frac{1}{2}y=0,05\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}x=0,025\\y=0,05\end{matrix}\right.\)

m muối =m NaHCO3+ mNa2CO3

\(=0,025\left(23+1+12+16.3\right)+0,025.\left(23.2+12+16.3\right)=4,75\left(g\right)\)