MNa2CO3=2ANa+AC+3AO=2*23+12+3*16=106g/mol
MHCl=AH+ACl=1+35,5=36,5g/mol
n=m/M=10,6/106=0,1 moli Na2CO3
n=m/M=4,25/36,5=0,116438 moli HCl
Este acid in exces =>0,116438-0,1=0,0164... moli exces
Na2CO3 + 2HCl -> 2NaCl + CO2↑ + H2O
1 mol Na2CO3.........................22,4 L CO2
0,1 moli Na2CO3.....................x L CO2
x=22,4*0,1=