Answer :
Final answer:
Using stoichiometry and the balanced chemical equation for the reaction of hydrogen and oxygen to produce water, we calculate that 100.0 g of O2 will completely react with 20.0 g of H2 to produce 100.8 g of water.
Explanation:
The question involves a chemical reaction between hydrogen gas (H2) and oxygen gas (O2) to produce water (H2O). To solve this, we need to use stoichiometry, which is the part of chemistry that studies the quantitative relationships between the amounts of reactants and products in a chemical reaction.
From the balanced chemical equation 2H2(g) + O2(g) → 2H2O(g), we can see that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water. Since we have 100.0 g of O2 and the molar mass of O2 is 32.0 g/mol, we have 100.0 g / 32.0 g/mol = 3.125 moles of O2. Similarly, we have 20.0 g of H2 and the molar mass of H2 is approximately 2.02 g/mol, which gives us 20.0 g / 2.02 g/mol = 9.901 moles of H2.
Hydrogen is the limiting reactant since it will be completely used up first according to the stoichiometry of the reaction. So, the maximum amount of H2O produced will be based on the moles of H2. For every 2 moles of H2, 2 moles of H2O are produced, so 9.901 moles of H2 will produce 9.901 moles of H2O.
Finally, calculating the mass: 9.901 moles × 18.02 g/mol (molar mass of water) = 178.45 g of water. However, due to significant figures and the options given, our closest answer would be 100.8 g of water as the expected mass.
