Answer:
0.45 grams of hydrogen and 3.53 grams of oxygen will be formed, when 4.05 grams of water is used.
Explanation:
The given reaction is [tex]2H_20 \rightarrow 2H_2 + O_2[/tex]
2 moles of [tex]H_2O[/tex] is used to form 2 moles of [tex]H_2[/tex] and 1 mole of [tex]O_2[/tex]
As 1 mole of [tex]H_2O=18 g[/tex], 1 mole of [tex]H_2=2 g[/tex], and 1 mole of [tex]O_2=32 g[/tex].
So, 36 grams of [tex]H_2O[/tex] is used to form 4 grams of [tex]H_2[/tex] and 32 grams of [tex]O_2[/tex]
Therefore, 1 gram of [tex]H_2O[/tex] is used to form 4/36 grams of [tex]H_2[/tex] and 32/36 grams of [tex]O_2[/tex]
Therefore, 4.05 gram of [tex]H_2O[/tex] is used to form (1/9)4.09=0.45 grams of [tex]H_2[/tex] and (8/9)4.05=3.53 grams of [tex]O_2[/tex].
Hence, 0.45 grams of hydrogen and 3.53 grams of oxygen will be formed, when 4.05 grams of water is used.
