Question: Given the balanced equation representing a reaction:
2H2 + O2 ==>2H2O
What is the total mass of water formed when 8 grams of hydrogen reacts completely with 64 grams of oxygen?
(1) 18 g (3) 56 g
(2) 36 g (4) 72 g
Answer:
"72 gram" is the total mass of water formed when 8 grams of hydrogen reacts completely with 64 grams of oxygen
Explanation:
We will do stoichiometry in this kind of problem:
[tex]8g of H_2O\times(\frac{1molO_2}{2.0g H_2})(\frac{2 mol H_2O}{2mol H_2})(\frac{18g H_2O}{1molH_2O}) = 72g H_2O[/tex]
[tex]64 g O_2 \times (\frac{1molO_2}{32g O_2})(\frac{2 mol H_2O}{1mol O_2})(\frac{16gH_2O}{1 mol H_2O}) = 72g H_2O[/tex]
Therefore, the total mass of water formed is 72 g.
The complete reaction of hydrogen and oxygen forms 72 grams of water.
The balanced chemical equation for the reaction has been:
[tex]\rm H_2\;+\;O_2\;\rightarrow\;2\;H_2O[/tex]
According to the balanced chemical equation,
1 moles of hydrogen reacts with 1 mole of oxygen, to give 2 moles of water.
Thus,
1 mole hydrogen requires 1 mole of oxygen
Moles can be given by:
[tex]\rm Moles\;=\;\dfrac{Mass}{Molecular\;mass}[/tex]
The moles of 8 grams hydrogen has been:
[tex]\rm Moles\;of \;hydrogen\;=\;\dfrac{8\;g}{2\;g/mol}[/tex]
Moles of hydrogen = 4 mol
The moles of 64 grams of oxygen has been:
[tex]\rm Moles\;of \;oxygen\;=\;\dfrac{64\;g}{32\;g/mol}[/tex]
Moles of oxygen = 2 mol
Since, there has been 2 moles of oxygen, for 4 moles of hydrogen, oxygen has been the limiting reagent.
From the balanced chemical equation:
1 mole oxygen = 2 moles water
2 mole oxygen = 2 × 2 moles water
2 mole oxygen = 4 moles water
The mass of 4 moles water can be given as:
Mass = Moles × Molecular mass
Mass of water = 4 mol × 18 g/mol
Mass of water = 72 g
The mass of water formed by complete reaction of 8 grams hydrogen has been 72 grams.
For more information about mass formed, refer to the link:
https://brainly.com/question/7481683
