Question

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Answer 1

\(n_{H_2}=\dfrac{2,8}{2}=1,4\left(mol\right)\)

\(2H_2+O_2\xrightarrow[]{t^{\circ}}2H_2O\)

2,8       →         2,8

\(\Rightarrow m_{H_2O}=2,8\cdot18=50,4\left(g\right)\)

\(\Rightarrow V_{H_2O}=\dfrac{m_{H_2O}}{D_{H_2O}}=\dfrac{50,4}{1}=50,4\left(ml\right)\)