\(n_C=\frac{m}{M}=\frac{24}{12}=2\left(mol\right)\)
\(n_{O_2}=\frac{V}{22,4}=\frac{11,2}{22,4}=0,5\left(mol\right)\)
PTHH :
\(C+O_2\rightarrow CO_2\)
Ta có : \(n_C:n_{O_2}=\frac{2}{1}:\frac{0,5}{1}=2:0,5\)
\(\Leftrightarrow C\) dư , \(O_2\) hết
\(n_Cpư=n_{O_2}=0,5\left(mol\right)\)
\(\Leftrightarrow n_C\) dư \(=2-0,5=1,5\left(mol\right)\)
\(\Leftrightarrow m_Cdư=n.M=12.1,5=18\left(g\right)\)
b/ \(n_{CO2}=n_{O2}=0,5\left(mol\right)\)
\(\Leftrightarrow m_{CO2}=n.M=0,5.44=22\left(g\right)\)
Vậy...
------Bài làm-----------
nC=mM=2412=2(mol)
nO2=V22,4=11,222,4=0,5(mol)
PTHH :
C+O2→CO2
Ta có : nC:nO2=21:0,51=2:0,5
⇔Cdư , O2 hết
nCpư=nO2=0,5(mol)
⇔nC dư =2−0,5=1,5(mol)
⇔mCdư=n.M=12.1,5=18(g)
b/ nCO2=nO2=0,5(mol)
⇔mCO2=n.M=0,5.44=22(g)
Vậy làm nha bạn !
