In the explosion of a hydrogen filled balloon, 0.50 g of hydrogen reacted with 4.0 g of oxygen to form how many grams of water vapor? (water vapor is the only product)
4.5g
Do the following:
1) Balance the following:
H2 + O2 -> H2O
2) Find the limiting reagent. The reactant (H2 or O2?) that produces the lesser amount of the product (H2O).
http://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm
To find the number of grams of water vapor formed in the explosion of the hydrogen-filled balloon, we first need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed, thus determining the maximum amount of product that can be formed.
To do this, we can compare the amount of hydrogen and oxygen used in the reaction. The balanced chemical equation for the reaction 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.
First, we need to convert the mass of hydrogen (0.50 g) and oxygen (4.0 g) into moles. We can use the molar mass of each element to make this conversion.
The molar mass of hydrogen (H₂) is 2 g/mol, and the molar mass of oxygen (O₂) is 32 g/mol.
Moles of hydrogen = mass of hydrogen / molar mass of hydrogen
= 0.50 g / 2 g/mol
= 0.25 mol
Moles of oxygen = mass of oxygen / molar mass of oxygen
= 4.0 g / 32 g/mol
= 0.125 mol
Now, we need to compare the number of moles of hydrogen and oxygen to determine the limiting reactant.
From the balanced equation, we know that the mole ratio of hydrogen to oxygen is 2:1. Therefore, for every 2 moles of hydrogen, we need 1 mole of oxygen.
For the amount of oxygen (0.125 mol) we have, we can calculate the number of moles required for the same amount of hydrogen.
Moles of oxygen required = (0.125 mol of hydrogen) / (2 mol of hydrogen/1 mol of oxygen)
= (0.125 mol * 1 mol of oxygen/2 mol of hydrogen)
= 0.0625 mol
Since we only have 0.125 mol of oxygen available, which is greater than the 0.0625 mol required, we can conclude that oxygen is the limiting reactant in this reaction.
Now, we can calculate the number of moles of water formed using the stoichiometry of the balanced equation.
From the balanced equation, we know that 2 moles of water are formed for every 1 mole of oxygen reacted.
Moles of water = moles of oxygen * (2 mol of water/1 mol of oxygen)
= 0.125 mol * 2 mol of water/1 mol of oxygen
= 0.25 mol
Finally, we can convert the number of moles of water to grams using the molar mass of water, which is 18 g/mol.
Mass of water vapor = moles of water * molar mass of water
= 0.25 mol * 18 g/mol
= 4.5 g
Therefore, 4.5 grams of water vapor will be formed in the explosion of the hydrogen-filled balloon.