\(n_{NaOH}=0,5.1\left(mol\right)\)
\(n_{hh\left(khi\right)}=1\left(mol\right)\Rightarrow n_{CO2}< 1\)
Vậy nNaOH/nCO2 < 1 vậy chất rắn khan gồm NaOH dư và Na2CO3
Gọi nCO2 = a mol
\(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)
2a_________a ______ a _______ (mol)
\(\Rightarrow\left\{{}\begin{matrix}n_{Na2CO3}=a\left(mol\right)\\n_{NaOH\left(dư\right)}=1-2a\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{cr\left(khan\right)}=m_{NaOH}+m_{Na2CO3}=\left(1-2a\right).40+106a=50,4\)
\(\Rightarrow a=0,4\left(mol\right)\)
\(n_{H2}=n_{hh}-n_{CO2}=1-0,4=0,6\left(mol\right)\)
Ta có PTHH :
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,6 ____________________0,6 (mol)
\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)
0,4______________________ 0,4 _____(mol)
\(\Rightarrow m=m_{Mg}+m_{MgCO3}=0,5.24+0,4.84=48\left(g\right)\)
