Question

Answer 1

Na2CO3+2HCl->2NaCl+H2O+CO2

0,1-----------0,2---------------------------0,1

n Na2CO3=\(\dfrac{10,6}{106}\)=0,1 mol

=>VCO2=0,1.22,4=2,24l

=>m HCl=0,2.36,5=7,3g

Answer 2

\(a,PTHH:Na_2CO_3+2HCl\rightarrow2NaCl+CO_2\uparrow+H_2O\)

\(b,n_{Na_2CO_3}=\dfrac{m}{M}=\dfrac{10,6}{106}=0,1mol\)

\(\rightarrow n_{CO_2}=0,1mol\) \(\Rightarrow V_{CO_2}=n.22,4=0,1.22,4=2,24l\)

\(\rightarrow n_{HCl}=0,2mol\) \(\Rightarrow m_{HCl}=n.M=0,2.36,5=7,3g\)