\(n_{BaCO_3}=\dfrac{29,55}{197}=0,15\left(mol\right)\)
\(n_{CO_2}=\dfrac{2,688}{22,4}=0,12\left(mol\right)\)
Theo bài ra: \(BaCO_3\) dư
PTHH: \(BaCO_3\underrightarrow{t^0}BaO+CO_2\)
Theo PTHH: \(n_{BaO}:n_{BaCO_3}=1:1\Rightarrow n_{BaO}=n_{BaCO_3}=0,15\left(mol\right)\)
\(\Rightarrow m_{BaO}=0,15.153=22,95\left(g\right)\)
\(\Rightarrow m_{BaCO_3\left(dư\right)}=\left(0,15-0,12\right).197=5,91\left(g\right)\)
\(n_{BaCO_3}=\dfrac{29,55}{197}=0,15\left(mol\right)\)
\(n_{CO_2}=\dfrac{2,688}{22,4}=0,12\left(mol\right)\)
PTHH: \(BaCO_3-t^o->BaO+CO_2\)
Theo PT ta lập tỉ lệ:
\(\dfrac{0,15}{1}>\dfrac{0,12}{1}\Rightarrow BaCO_3\) dư. CO2 hết => tính theo \(n_{CO_2}\)
Theo PT: \(n_{BaO}=n_{CO_2}=0,12\left(mol\right)\)
=> \(m_{BaO}=0,12.153=18,36\left(g\right)\)
Theo PT: \(n_{BaCO_3\left(pư\right)}=n_{CO_2}=0,12\left(mol\right)\)
=> \(n_{BaCO_3\left(dư\right)}=0,15-0,12=0,03\left(mol\right)\)
=> \(m_{BaCO_3\left(dư\right)}=0,03.197=5,91\left(g\right)\)
\(n_{BaCO_3}=\dfrac{29,55}{197}=0,15\left(mol\right)\)
\(n_{CO_2}=\dfrac{2,688}{22,4}=0,12\left(mol\right)\)
Theo đề bài: \(BaCO_3\) dư
PTHH:
\(BaCO_3\underrightarrow{t^0}BaO+CO_2\)
0,12...........0,12...........0,12
\(\Rightarrow m_{BaO}=0,12.153=18,36\left(g\right)\)
\(\Rightarrow m_{BaCO_3\left(dư\right)}=\left(0,15-0,12\right).197=5,91\left(g\right)\)
