In the explosion of a hydrogen-filled balloon, 0.54 g of hydrogen reacted with 6.6 g of oxygen to form how many grams of water vapor?

Figure the moles of each.

.54/2=.27 moles H2
6.6/32=.21moles O2
Balance the reaction:

2H2+O2>>2H2O

well, you need twice as much H2 as O2, so H2 is the limiting reactant. Use the limiting reactant.

.27molesH2 (2moleswater/2molesH2)=.27moleswater

figure the mass of .27 moles water

To determine the grams of water vapor formed in the explosion, you need to use the concept of stoichiometry. Stoichiometry involves using balanced chemical equations to relate the amounts of reactants and products in a chemical reaction.

First, you need to write the balanced chemical equation for the reaction between hydrogen and oxygen to form water:

2 H2 + O2 -> 2 H2O

According to the equation, 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.

Now, let's calculate the moles of hydrogen and oxygen in the given quantities:

Moles of hydrogen = mass of hydrogen / molar mass of hydrogen
= 0.54 g / 2.016 g/mol (molar mass of hydrogen)
≈ 0.268 moles

Moles of oxygen = mass of oxygen / molar mass of oxygen
= 6.6 g / 31.998 g/mol (molar mass of oxygen)
≈ 0.206 moles

Based on the balanced equation, we can see that the reaction requires 2 moles of hydrogen for every 1 mole of oxygen. However, since the moles of oxygen are less than half the moles of hydrogen, we consider that oxygen is the limiting reactant.

To determine the moles of water vapor formed, we apply the stoichiometric ratio from the balanced equation:

Moles of water vapor = moles of oxygen (since oxygen is the limiting reactant)
= 0.206 moles

Finally, we can convert moles of water vapor to grams:

Mass of water vapor = moles of water vapor * molar mass of water
= 0.206 moles * 18.015 g/mol (molar mass of water)
≈ 3.71 g

Therefore, approximately 3.71 grams of water vapor would be formed in the explosion of the hydrogen-filled balloon.