How many grams of water are produced if we react 3 moles of hydrogen with 3 moles of oxygen?
About 60 grams
About 40 grams
About 20 grams
About 80 grams​

How many grams of water are produced if we react 3 moles of hydrogen with 3 moles of oxygen About 60 grams About 40 grams About 20 grams About 80 grams class=

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Answer:

About 60 grams

Explanation:

The balanced equation of the reaction is given as;

2H2(g) + O2(g) → 2H2O(l)

From the equation;

2 mol of H2 reacts with 1 mol of O2

3 mol of H2 would require 3/2 mol of O2 (considering the 2:1 ratio)

The limiting reactant is H2 as it would be used up before H2. It determines the amount of product that would be formed.

2 mol of H2 produces 2 mol of H2O

3 mol of H2 would produce 3 mol of H2O (Considering the 1 : 1 ratio)

Converting moles to mass;

Mass = Molar mass * Number of moles

Mass = 18 * 3 = 54 g

The correct option is; About 60 grams

Taking into account the reaction stoichiometry, the correct answer is the first answer: about 60 grams of H₂O are formed when 3 moles of hydrogen react with 3 moles of oxygen.

In first place, the balanced reaction is:

2 H₂ + O₂ → 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • H₂: 2 moles
  • O₂: 1 mole
  • H₂O: 2 moles

The molar mass of the compounds is:

  • H₂: 1 g/mole
  • O₂: 32 g/mole  
  • H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • H₂: 2 moles× 1 g/mole= 2 grams
  • O₂: 1 mole× 32 g/mole= 32 grams  
  • H₂O: 2 moles× 18 g/mole= 36 grams

Limiting reagent

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction and a simple rule of three as follows: if by stoichiometry 1 mole of O₂ reacts with 2 moles of H₂,  if 3 moles of O₂ react how much moles of H₂ will be needed?

[tex]amount of moles of H_{2} =\frac{3 moles of O_{2}x2 moles of H_{2} }{1 mole of O_{2} }[/tex]

moles of H₂= 6 moles

But 6 moles of H₂ are not available, 3 moles are available. Since you have less moles than you need to react with 3 moles of O₂, H₂ will be the limiting reagent.

Mass of water produced

The following rule of three can be applied, considering the stoichiometry ans the limiting reagent: if by reaction stoichiometry 2 moles of H₂ form 36 grams of H₂O, 3 moles of H₂ form how much mass of H₂O?

[tex]mass of H_{2}O=\frac{3 moles of H_{2}x36 grams of H_{2}O }{2 moles of H_{2}}[/tex]

mass of H₂O= 54 grams

Then, 54 grams of H₂O are formed when 3 moles of hydrogen react with 3 moles of oxygen.

Finally, the correct answer is the first answer: about 60 grams of H₂O are formed when 3 moles of hydrogen react with 3 moles of oxygen.

Learn more about reaction stoichiometry:

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