\(2H_2+O_2\underrightarrow{t^o}2H_2O\)
0,25 0,125 0,25
\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{8,4}{22,4}=0,375mol\)
\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{2,8}{22,4}=0,125mol\)
so sánh tỉ lệ mol:
\(\dfrac{0,375}{2}>\dfrac{0,125}{1}\)
\(m_{H_2O}=n.M=0,25.18=4,5g\)
2H2 + O2 -to-➢ 2H2O
\(n_{H_2}=\dfrac{8,4}{22,4}=0,375\left(mol\right)\)
\(n_{O_2}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\)
Theo PT: \(n_{H_2}=2n_{O_2}\)
Theo bài: \(n_{H_2}=3n_{O_2}\)
Vì \(3>2\) ⇒ H2 dư, O2 hết
Theo PT: \(n_{H_2O}=2n_{O_2}=2\times0,125=0,25\left(mol\right)\)
\(\Rightarrow m_{H_2O}=0,25\times18=4,5\left(g\right)\)
