Question

what is the total mass of water formed when 8 grams of hydrogen completely reacts with 64 grams oxygen

Answer 1

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:

$8g of H_2O\times(\frac{1molO_2}{2.0g H_2})(\frac{2 mol H_2O}{2mol H_2})(\frac{18g H_2O}{1molH_2O}) = 72g H_2O$

$64 g O_2 \times (\frac{1molO_2}{32g O_2})(\frac{2 mol H_2O}{1mol O_2})(\frac{16gH_2O}{1 mol H_2O}) = 72g H_2O$

Therefore, the total mass of water formed is 72 g.

Answer 2

The complete reaction of hydrogen and oxygen forms 72 grams of water.

The balanced chemical equation for the reaction has been:

$\rm H_2\;+\;O_2\;\rightarrow\;2\;H_2O$

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:

$\rm Moles\;=\;\dfrac{Mass}{Molecular\;mass}$

The moles of 8 grams hydrogen has been:

$\rm Moles\;of \;hydrogen\;=\;\dfrac{8\;g}{2\;g/mol}$

Moles of hydrogen = 4 mol

The moles of 64 grams of oxygen has been:

$\rm Moles\;of \;oxygen\;=\;\dfrac{64\;g}{32\;g/mol}$

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