Question

Answer 1

a, Ta có: \(n_{CO_2}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\)

\(n_{NaOH}=0,6.2=1,2\left(mol\right)\)

\(\Rightarrow\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{1,2}{0,75}=1,6\) 

⇒ Vậy pư tạo 2 muối Na₂CO₃ và NaHCO₃.

Gọi: \(\left\{{}\begin{matrix}n_{Na_2CO_3}=a\left(mol\right)\\n_{NaHCO_3}=b\left(mol\right)\end{matrix}\right.\)

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

_____a______2a_________a (mol)

\(CO_2+NaOH\rightarrow NaHCO_3\)

__b______b________b (mol)

\(\Rightarrow\left\{{}\begin{matrix}a+b=0,75\\2a+b=1,2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=0,45\left(mol\right)\\b=0,3\left(mol\right)\end{matrix}\right.\)

⇒ m _muối = m_Na2CO3 + m_NaHCO3 = 0,45.106 + 0,3.84 = 72,9 (g)

b, PT: \(Na_2CO_3+BaCl_2\rightarrow2NaCl+BaCO_{3\downarrow}\)

________0,45______________________0,45 (mol)

\(\Rightarrow m_{BaCO_3}=0,45.197=88,65\left(g\right)\)