Answer:
The mass of the product is 73,06 g
Explanation:
The reaction to form water is:
2H2 + O2 ---> 2H2O
35 g 65 g
Molar mass H2 = 2g/m ---> I have 17,5 moles (35g / 2 g/m)
Molar mass O2 = 32g/m ---> I have 2,03 moles (65g / 32g/m)
2 moles of H2 __reacts with __ 1 mol O2
17,5 moles of H2 __ reacts with ___ (17,5 m . 1 m) / 2 m = 8,75 m
I have just 2,03 m of O2 and I need 8,75 m so O2 is my limiting reagent
1 mol of O2 ___ reacts with 2 moles H2
2,03 moles of O2 __ reacts with (2,03 m . 2 m) / 1 m = 4,06 m
I have 17,5 moles of H2 and I need 4,06 m. H2 is my excess reagent
<u><em>"All operations are done with the limiting reagent"</em></u>
1 mol O2 are required____ to form 2 H2O
2,03 mol O2 are required __ to form (2,03m . 2m) / 1m = 4,06 m
Molar mass of water: 18 g/m
Mass of water ----> Molar mass . Moles water = 18 g/m . 4,06 m = 73,08g
