Calculation of the mass of water formed from the reaction of hydrogen and oxygen:

Given data:
- Mass of hydrogen = 4 g
- Mass of oxygen = 20 g

Step 1: Write the balanced chemical equation for the reaction of hydrogen and oxygen to form water:

2H2 + O2 → 2H2O

Step 2: Determine the limiting reactant:

- Calculate the number of moles of hydrogen and oxygen using their respective molar masses:
- Molar mass of hydrogen = 1 g/mol
- Molar mass of oxygen = 16 g/mol

- Moles of hydrogen = 4 g ÷ 1 g/mol = 4 mol
- Moles of oxygen = 20 g ÷ 16 g/mol = 1.25 mol

- The reactant that is completely consumed in the reaction is the limiting reactant. In this case, oxygen is the limiting reactant because it is present in lesser quantity than required for the reaction.

Step 3: Calculate the mass of water formed:

- From the balanced equation, we can see that 1 mole of oxygen reacts with 2 moles of hydrogen to form 2 moles of water.
- Moles of oxygen used in the reaction = 1.25 mol
- Moles of water formed = 2 × 1.25 mol = 2.5 mol

- Mass of water formed = moles of water × molar mass of water
- Molar mass of water = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
- Mass of water formed = 2.5 mol × 18.02 g/mol = 45.05 g

Therefore, the mass of water formed from the reaction of 4 g of hydrogen and 20 g of oxygen is 45.05 g.

Explanation: When hydrogen and oxygen react to form water, they undergo a chemical reaction that involves the breaking and formation of chemical bonds. The balanced chemical equation for this reaction shows that 2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water. The given data provides us with the masses of hydrogen and oxygen, which can be converted to moles using their respective molar masses. The limiting reactant is determined by comparing the number of moles of each reactant and identifying the one that is present in lesser quantity than required for the reaction. The mass of water formed is then calculated using stoichiometry and the molar mass of water.

22.5g