a, \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
\(n_{O_2}=\dfrac{8,4}{22,4}=0,375\left(mol\right)\)
PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
Xét tỉ lệ: \(\dfrac{0,25}{2}< \dfrac{0,375}{1}\), ta được O2 dư.
Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{H_2}=0,125\left(mol\right)\Rightarrow n_{O_2\left(dư\right)}=0,375-0,125=0,25\left(mol\right)\)
\(\Rightarrow V_{O_2\left(dư\right)}=0,25.22,4=5,6\left(l\right)\)
b, Theo PT: \(n_{H_2O}=n_{H_2}=0,25\left(mol\right)\Rightarrow m_{H_2O}=0,25.18=4,5\left(g\right)\)
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