PTHH: \(CH_4+2O_2-^{t^o}\rightarrow CO_2+2H_2O\)
Sản phẩm tạo thành : CO2 và H2O
\(n_{CH_4}=\frac{16,8}{22,4}=0,75\left(mol\right)\)
Theo PT: \(n_{CO_2}=n_{CH_4}=0,75\left(mol\right)\)
\(n_{H_2O}=2n_{CH_4}=1,5\left(mol\right)\)
\(\Rightarrow m_{CO_2}=0,75.44=33\left(g\right)\)
\(m_{H_2O}=1,5.18=27\left(g\right)\)
Theo PT: \(n_{O_2\left(pứ\right)}=2n_{CH_4}=1,5\left(mol\right)\)
\(\Rightarrow V_{O_2\left(pứ\right)}=1,5.22,4=33,6\left(l\right)\)
Vì ban đầu lấy dư 20% O2
\(\Rightarrow V_{O_2\left(bđ\right)}=33,6.120\%=40,32\left(l\right)\)
a,\(PTHH:CH_4+2O_2\rightarrow2H_2O+CO_2\)
Theo PT: ___1 ___ 2 ____ 2______ 1
Theo đề: ___0,75___1,5 ___ 1,5___0,75 _
b, \(n_{CH4}=\frac{16,8}{22,4}=0,75\left(mol\right)\)
\(m_{H2O}=1,5.18=27\left(g\right)\)
\(\Rightarrow V_{O2}=1,5.22,4=33,6\left(l\right)\)
