a, PT: BaCO3+2HCl→BaCl2+CO2+H2OBaCO3+2HCl→BaCl2+CO2+H2O
Ta có: nCO2=0,44822,4=0,02(mol)nCO2=0,44822,4=0,02(mol)
Theo PT: nHCl=2nCO2=0,04(mol)nHCl=2nCO2=0,04(mol)
⇒CMHCl=0,040,1=0,4M⇒CMHCl=0,040,1=0,4M
b, Muối thu được sau pư gồm: BaCl2 và NaCl.
Theo PT: nBaCO3=nBaCl2=nCO2=0,02(mol)nBaCO3=nBaCl2=nCO2=0,02(mol)
⇒mBaCO3=0,02.197=3,94(g)⇒mBaCO3=0,02.197=3,94(g)
⇒mNaCl=5−3,94=1,06(g)⇒mNaCl=5−3,94=1,06(g)
⇒ m muối thu được = mNaCl + mBaCl2 = 1,06 + 0,02.208 = 5,22 (g)
Bạn tham khảo nhé!
a) PTHH: \(BaCO_3+2HCl\rightarrow BaCl_2+H_2O+CO_2\uparrow\)
Ta có: \(n_{CO_2}=\frac{0,448}{22,4}=0,02\left(mol\right)\)
\(\Rightarrow n_{HCl}=0,04mol\) \(\Rightarrow C_{M_{HCl}}=\frac{0,04}{0,01}=4\left(M\right)\)
b) Theo PTHH: \(n_{CO_2}=n_{BaCO_3}=n_{BaCl_2}=0,02mol\)
\(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=5-0,02\cdot197=1,06\left(g\right)\\m_{BaCl_2}=0,02\cdot208=4,16\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m_{muối}=1,06+4,16=5,22\left(g\right)\)