In the explosion of a hydrogen-filled balloon, 0.80g of hydrogen reacted with 6.4g of oxygen. How many grams of water vapor are formed? (Water vapor is the only product.)
Express your answer using two significant figures.
This is a limiting reagent problem(except both are limiting reagent. :-).
2H2 + O2 ==> 2H2O
Convert 0.80g to mols. mol = grams/molar mass.
Convert 6.4 g O2 to mols. Same process.
Using the coefficient in the balanced equation, convert mols H2 to mols H2O.
Do the same for mols O2 to mols H2O.
Since mols H2O are the same for both reactants there is no limiting reagent and both reagent will be used completely.
Then grams H2O = mols of either x molar mass of that reactant. You can do each of them and show that grams H2O is the same no matter which is used.
To determine the number of grams of water vapor formed, we need to use the balanced chemical equation for the reaction between hydrogen (H₂) and oxygen (O₂) to form water (H₂O).
The balanced chemical equation is:
2H₂ + O₂ → 2H₂O
From the equation, we can see that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.
First, let's calculate the number of moles of hydrogen and oxygen involved in the reaction:
The molar mass of hydrogen (H₂) is 2.016g/mol. We can calculate the number of moles of hydrogen using the given mass of hydrogen:
moles of hydrogen = mass of hydrogen / molar mass of hydrogen
= 0.80g / 2.016g/mol
= 0.397 mol (rounded to three decimal places)
Next, let's calculate the number of moles of oxygen:
The molar mass of oxygen (O₂) is 32g/mol. We can calculate the number of moles of oxygen using the given mass of oxygen:
moles of oxygen = mass of oxygen / molar mass of oxygen
= 6.4g / 32g/mol
= 0.200 mol (rounded to three decimal places)
Now, let's use the mole ratios from the balanced chemical equation to determine the number of moles of water formed. 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water. Therefore, the molar ratio of hydrogen to water is 2:2, or 1:1. That means the number of moles of water formed is equal to the number of moles of hydrogen.
moles of water = moles of hydrogen = 0.397 mol
Finally, let's convert the moles of water to grams by multiplying by the molar mass of water:
mass of water = moles of water × molar mass of water
= 0.397 mol × 18.015 g/mol
= 7.14 g (rounded to two significant figures)
Therefore, the number of grams of water vapor formed in the reaction is 7.14 g.
To determine the grams of water vapor formed in the explosion, we need to use the balanced chemical equation for the reaction between hydrogen and oxygen to form water.
The balanced equation is:
2H2 + O2 -> 2H2O
From the equation, we can see that 2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water.
To find the number of moles of hydrogen and oxygen in the given masses, we will use the molar masses. The molar masses are:
Hydrogen (H2) = 2g/mol
Oxygen (O2) = 32g/mol
First, we calculate the number of moles of hydrogen by dividing the mass of hydrogen by its molar mass:
Number of moles of hydrogen = 0.80g / (2 g/mol) = 0.40 mol
Next, we calculate the number of moles of oxygen using the same method:
Number of moles of oxygen = 6.4g / (32 g/mol) = 0.20 mol
Since the reaction occurs according to the stoichiometry of 2:1 for hydrogen to oxygen and 2:2 for hydrogen to water, we can see that the limiting reactant is oxygen.
Therefore, we can conclude that 0.20 moles of oxygen reacts to form 0.20 moles of water vapor.
Finally, we convert moles of water vapor to grams by multiplying the number of moles by the molar mass of water:
Molar mass of water (H2O) = 18g/mol
Mass of water vapor = 0.20 mol x (18 g/mol) = 3.6 g
Thus, the number of grams of water vapor formed in the explosion is 3.6 g.