$$2H_2(g) + O_2(g) → 2H_2O(g)$$
The molar mass of hydrogen ($H_2$) is 2.016 g/mol, and the molar mass of oxygen ($O_2$) is 32.00 g/mol. The molar mass of water ($H_2O$) is 18.02 g/mol.
To determine the limiting reactant, we need to compare the number of moles of hydrogen and oxygen available for the reaction.
First, we convert the given masses of hydrogen and oxygen to moles:
$$moles\space of \space H_2 = \frac{0.90 \space g}{2.016 \space g/mol} = 0.447 \space mol$$
$$moles\space of \space O_2 = \frac{7.2 \space g}{32.00 \space g/mol} = 0.225 \space mol$$
Comparing the number of moles of hydrogen and oxygen, we find that hydrogen is the limiting reactant because it is present in a smaller amount compared to oxygen.
Therefore, all of the hydrogen will react, and the amount of water vapor produced will be determined by the amount of hydrogen available.
According to the balanced chemical equation, 2 moles of hydrogen produce 2 moles of water. So, 0.447 moles of hydrogen will produce:
$$moles\space of \space H_2O = 0.447 \space mol \spaceH_2 \times \frac{2 \space mol \space H_2O}{2 \space mol \space H_2} = 0.447 \space mol \space H_2O$$
Finally, we convert the moles of water vapor back to grams:
$$mass\space of \space H_2O = 0.447 \space mol \times 18.02 \space g/mol = 8.05 \space g$$
Therefore, 8.05 grams of water vapor will be formed in the explosion of the hydrogen-filled balloon.
