In the explosion of a hydrogen-filled balloon, 0.30g of hydrogen reacted with 2.4g of oxygen to form how many grams of water vapor? (Water vapor is the only product.)

2.7g

To determine the grams of water vapor formed in the reaction, we need to find the limiting reactant first.

1. Convert the masses of hydrogen and oxygen to moles using their molar masses:
- Hydrogen (H2): 1 g/mol
- Oxygen (O2): 32 g/mol

Moles of hydrogen: 0.30 g / 2 g/mol = 0.15 mol
Moles of oxygen: 2.4 g / 32 g/mol = 0.075 mol

2. To find the limiting reactant, compare the moles of hydrogen and oxygen. The reactant that has the smallest mole ratio is the limiting reactant. In this case, oxygen has a smaller mole ratio:
Hydrogen:Oxygen mole ratio = 0.15 mol / 0.075 mol = 2:1

3. Now that we know oxygen is the limiting reactant, we can calculate the moles of water vapor formed. The balanced chemical equation is:
2H2 + O2 -> 2H2O

Since the mole ratio between oxygen and water vapor is 1:2, the moles of water vapor formed is also 0.075 mol.

4. Finally, we can find the mass of the water vapor by multiplying the number of moles by the molar mass of water (H2O):
Molar mass of H2O = 18 g/mol

Mass of water vapor formed = 0.075 mol * 18 g/mol = 1.35 g

Therefore, 1.35 grams of water vapor will be formed.

To find the number of grams of water vapor formed in the explosion, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.

First, we calculate the number of moles of hydrogen and oxygen:

Number of moles of hydrogen = mass of hydrogen / molar mass of hydrogen
Number of moles of hydrogen = 0.30g / 2.016g/mol = 0.149 moles of hydrogen

Number of moles of oxygen = mass of oxygen / molar mass of oxygen
Number of moles of oxygen = 2.4g / 32.00g/mol = 0.075 moles of oxygen

Next, we need to calculate the mole ratio between hydrogen and oxygen in the balanced chemical equation for the reaction. The balanced equation for the reaction between hydrogen and oxygen to form water is:

2H2 + O2 -> 2H2O

From the balanced equation, the ratio of hydrogen to oxygen is 2:1. This means that for every 2 moles of hydrogen, we need 1 mole of oxygen.

Since the ratio of hydrogen to oxygen is 2:1, we can see that the 0.075 moles of oxygen is the limiting reactant because it is less than half of the moles of hydrogen (0.149 moles).

Now, we can calculate the number of moles of water vapor formed using the limiting reactant:

Number of moles of water vapor = 0.075 moles of oxygen x (2 moles of water vapor / 1 mole of oxygen)
Number of moles of water vapor = 0.15 moles of water vapor

Finally, we calculate the mass of water vapor formed:

Mass of water vapor = number of moles of water vapor x molar mass of water
Mass of water vapor = 0.15 moles x 18.015g/mol = 2.7025g

Therefore, the grams of water vapor formed in the explosion is approximately 2.70g.

45 g