a) mgiảm = mCO2 = 12 - 7,6 = 4,4 (g)
=> \(n_{CO_2}=\dfrac{4,4}{44}=0,1\left(mol\right)\)
PTHH: \(CaCO_3\xrightarrow[]{t^o}CaO+CO_2\uparrow\)
0,1<-------0,1<-----0,1
\(n_{CaCO_3\left(bđ\right)}=\dfrac{12}{100}=0,12\left(mol\right)\Rightarrow n_{CaCO_3\left(dư\right)}=0,12-0,1=0,02\left(mol\right)\)
=> \(\left\{{}\begin{matrix}\%m_{CaCO_3}=\dfrac{0,02.100}{7,6}.100\%=26,32\%\\\%m_{CaO}=100\%-26,32\%=73,68\%\end{matrix}\right.\)
b) PTHH: \(CaO+2HCl\rightarrow CaCl_2+H_2O\)
\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
0,02----------------------------->0,02
\(n_{NaOH}=0,125.0,2=0,025\left(mol\right)\)
Xét \(T=\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,025}{0,02}=1,25\) => Pư tạo 2 muối
PTHH: \(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)
0,025--->0,0125-->0,0125
\(Na_2CO_3+CO_2+H_2O\rightarrow2NaHCO_3\)
0,0075<--0,0075---------->0,015
=> \(\left\{{}\begin{matrix}C_{M\left(Na_2CO_3\right)}=\dfrac{0,005}{0,125}=0,04M\\C_{M\left(NaHCO_3\right)}=\dfrac{0,015}{0,0125}=0,12M\end{matrix}\right.\)
